A compound containing only carbon, hydrogen, and oxygen is analyzed using combustion analysis. When 50.1 g of the compound is burned, 91.8 g of carbon dioxide and 25.1 g of water are collected. In order to determine the moles of carbon in the compound, first determine the moles of carbon dioxide that were produced from the combustion.

Answers

Answer 1

Answer:

[tex]C_{3}H_4O_2[/tex]

Explanation:

Hello,

In this case, since the carbon of the initial compound is present in the carbon dioxide product, we can compute the mass and moles of carbon in the compound:

[tex]n_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =2.09molC\\\\m_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2}*\frac{12gC}{1molC} =25.0gC[/tex]

Next, the mass and moles of hydrogen in the compound, is contained in the yielded amount of water, thus, we compute the mass and moles of hydrogen in the compound:

[tex]n_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =2.79molH\\\\m_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} *\frac{1gH}{1molH} =2.79gH[/tex]

In such a way, the mass of oxygen comes from the mass of the compound minus the mass of carbon and oxygen:

[tex]m_O=50.1g-25.0g-2.79g=22.31gO[/tex]

And the moles:

[tex]n_O=22.31gO*\frac{1molO}{16gO}=1.39molO[/tex]

Then, we compute the subscripts by diving the moles of C, H and O by the moles of oxygen as the smallest moles:

[tex]C:\frac{2.09}{1.39}=1.5 \\\\H:\frac{2.79}{1.39}=2\\ \\O:\frac{1.39}{1.39} =1[/tex]

After that, we write:

[tex]C_{1.5}H_2O[/tex]

Which must be shown in whole number only, thereby we multiply the subscripts by 2, so the empirical formula turns out:

[tex]C_{3}H_4O_2[/tex]

Best regards.


Related Questions

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Rank the following elements in order from least to most number of moles of atoms in a 10.0 g sample: Sn, Si, Se, S

Answers

Answer:

[tex]\rm Sn[/tex], [tex]\rm Se[/tex], [tex]\rm S[/tex], [tex]\rm Si[/tex].

Explanation:

The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. This quantity can help estimate the number of moles of atoms in each of these four [tex]10.0\; \rm g[/tex] samples.  

Look up the relative atomic mass for each of these four elements (on a modern periodic table.)

[tex]\rm Si[/tex]: [tex]28.085[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm Se[/tex]: [tex]78.971[/tex].[tex]\rm Sn[/tex]: [tex]118.710[/tex].

The relative atomic mass of [tex]\rm Si[/tex] is (approximately) [tex]28.085[/tex]. Therefore, the each mole of silicon atoms would have a mass of approximately [tex]28.085\; \rm g[/tex]. How many moles of silicon atoms would there be in a [tex]10.0\; \rm g[/tex] sample?

Given:

[tex]m(\rm Si) = 10.0\; \rm g[/tex]. [tex]M(\mathrm{Si}) = 28.085\; \rm g \cdot mol^{-1}[/tex].

Number of mole of silicon atoms in the sample: [tex]\displaystyle n(\mathrm{Si}) = \frac{m(\mathrm{Si})}{M(\mathrm{Si})} = \frac{10.0\; \rm g}{28.085\; \rm g \cdot mol^{-1}}\approx 0.356\; \rm mol[/tex].

Similarly:

[tex]\displaystyle n(\mathrm{S}) = \frac{m(\mathrm{S})}{M(\mathrm{S})} = \frac{10.0\; \rm g}{32.06\; \rm g \cdot mol^{-1}}\approx 0.312\; \rm mol[/tex].

[tex]\displaystyle n(\mathrm{Se}) = \frac{m(\mathrm{Se})}{M(\mathrm{Se})} = \frac{10.0\; \rm g}{78.971\; \rm g \cdot mol^{-1}}\approx 0.127\; \rm mol[/tex].

[tex]\displaystyle n(\mathrm{Sn}) = \frac{m(\mathrm{Sn})}{M(\mathrm{Sn})} = \frac{10.0\; \rm g}{118.710\; \rm g \cdot mol^{-1}}\approx 0.0842\; \rm mol[/tex].

Therefore, among these [tex]10.0\; \rm g[/tex] samples:

[tex]n(\mathrm{Sn}) < n(\mathrm{Se}) < n(\mathrm{S}) < n(\mathrm{Si})[/tex].

It is not a coincidence that among these four samples, the one with the fewest number of atoms corresponds to the element with the largest relative atomic mass.

Consider two elements, with molar mass [tex]M_1[/tex] and [tex]M_2[/tex] each. Assume that [tex]n_1[/tex] moles and [tex]n_2[/tex] moles of atoms of each element were selected, such that the mass of both samples is [tex]m[/tex]. That is:

[tex]m = n_1\cdot M_1[/tex].

[tex]m = n_2\cdot M_2[/tex].

Equate the right-hand side of these two equations:

[tex]n_1 \cdot M_1 = n_2\cdot M_2[/tex].

[tex]\displaystyle \frac{n_1}{n_2} = \frac{M_2}{M_1} = \frac{1/M_1}{1/M_2}[/tex].

In other words, the number of moles atoms in two equal-mass samples of two elements is inversely proportional to the molar mass of the two elements (and hence inversely proportional to the formula mass of the two elements.) That explains why in this question, the sample containing the smallest number of atoms corresponds to element with the largest relative atomic mass among those four elements.

How to separate given mixture?

Answers

Answer:

Chromatography involves solvent separation on a solid medium.

Distillation takes advantage of differences in boiling points.

Evaporation removes a liquid from a solution to leave a solid material.

Filtration separates solids of different sizes.

Explanation:

4. CHALLENGE Suppose you had a mixture of sand and small,
hollow beads. How might you separate the mixture?

Answers

I'm not sure if this is the answer but maybe oil.

1) The speed constant for the second order reaction in the gas phase
It varies with the temperature according to the table below. Calculate the activation energy for the process, according to Arhhenius' equation

Answers

Answer:

41.7 kJ/mol

Explanation:

ln(k) = ln(A) − Eₐ/(RT)

Pick any two points.  I'll choose 100°C and 400°C.

When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:

ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)

When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:

ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)

Subtract the two equations and solve:

ln(4.40×10⁻⁷) −  ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)

5.991 = 0.00120 Eₐ/R

Eₐ/R = 5013.4

Eₐ = 41700 J/mol

Eₐ = 41.7 kJ/mol

The smallest form of matter that still retains the properties of an element

Answers

Answer:

atom

Explanation:

the atom is the smallest form.

Although there are smaller forms of matter (subatomic quarks, photons, electrons) an atom is the smallest that a form of matter could be that still possesses the properties of that element. An atom is classified into being a certain element based on how many protons it has.

What does chemical equations and chemical formulas have in common?​

Answers

Answer:

Chemical symbols refer to chemical elements only. They do not necessarily refer to atoms of that element, but also to ions.

Explanation:

Chemical symbols refer to chemical elements only

If 5.00g of iron metal is reacted with 0.950g of Cl2 gas, how many grams of ferric chloride (FeCl3) will form?

Answers

Answer:

1.45g of FeCl3

Explanation:

The equation of the reaction is given as;

2Fe + 3Cl2 --> 2FeCl3

2 mol of Fe reracts with 3 mol of Cl2 to form 2 mol of FeCl3

Upon converting to mass using;

Mass = Number of moles * Molar mass

( 2 * 55.85 = 111.7g ) of Fe reacts with ( 3 * 71 = 213g ) of Cl2 to form ( 2 * 162.2 = 324.4g) of FeCl3

Cl2 is the limiting reactant as it determines how much of FeCl3 is formed

213g of Cl2 = 324.4g of FeCl3

0.950g of Cl2 = x

x = (0.950 * 324.4 ) / 213

x = 1.45g of FeCl3

8 You are given 20.00g of a dry mixture of sand and table salt.
After adding water and filtering, you are left with wet sand on
the filter paper. The filter paper and sand is then dried and the
mass of the dry sand alone is found to be 5.00g. What was the
% sand in the original mixture?

Answers

Answer:

[tex]\%sand=25\%[/tex]

Explanation:

Hello.

In this case, given the mass of the mixture, we can define it in terms of the mass of sand and table salt as shown below:

[tex]m_{sand}+m_{salt}=20.00g[/tex]

Moreover, as after filtering, the mass of dry sand turns out 5.00 g, we can compute the % sand in the original mixture by dividing this value over the mass of the mixture as shown below:

[tex]\%sand=\frac{m_{sand}}{m_{mixture}}*100\%\\ \\\%sand=\frac{5.00g}{20.00g}* 100\%\\\\\%sand=25\%[/tex]

Best regards!

Select the term that matches each definition:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion.
b) The mass of a salt in grams that will dissolve in 100 mL of water.
c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved.
d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation.
e) The maximum number of moles of a salt that will dissolve in 1 L of solution.
*** Answer options for all questions: ***
- Solubility
- Molar Solubility
- Solubility product constant
- Common ion effect
- Saturated Solution

Answers

Answer:

a) common ion effect

b) solubility

c) saturated solution

d) solubility product constant

e) molar solubility

Explanation:

When a substance, say BA2 is dissolved in a solution and another substance CA2 is dissolved in the same solution. The solubility of BA2 is decreased due to the addition of CA2. This is known as common ion effect.

The mass of a substance that will dissolve in a given Volume of solvent is known as it's solubility.

The molar solubility is the amount of moles of solvent that dissolves in 1 dm^3 of solvent.

A solution that contains just as much solute as it can normally hold at a given temperature is known as a saturated solution.

Lastly, the product of molar solubilites raised to the power of the molar coefficient is know as the solubility product constant.

The correct matches and their explanation are:

a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion: Option 4. common ion effect

It relates to the equilibrium effect that occurs due to the addition of common ions.

b) The mass of salt in grams that will dissolve in 100 mL of water: Option 1. solubility

Solubility is the property of solute to form a solution with the solvent of another substance.

c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Option 5. saturated solution

When the solution cannot accommodate any more addition of solute of a substance is called a saturated solution.

d)  The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation: Option 3. solubility product constant

It is an equilibrium constant for the solid solute dissolved in the solution.

e) The maximum number of moles of a salt that will dissolve in 1 L of solution: Option 2. molar solubility

Before the saturation of a solution, the amount of solute it can dissolve is called molar solubility.

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3. A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and
found it read 196.5 9. What was the student's percent error?

Answers

Answer:

The answer is 1.71 %

Explanation:

The percentage error of a certain measurement can be found by using the formula

[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]

From the question

actual mass = 200 g

error = 200 - 196.59 = 3.41

We have

[tex]p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705[/tex]

We have the final answer as

1.71 %

Hope this helps you

A. Express each Fraction in Decimal form. Round off your answers into
thousandths place.
1. 3/4​

Answers

Answer:

0.75

Explanation:

3/4=0.75 so therefore its 0.75

Which neutral atom is isoelectronic with Cl-??

Answers

And we can see that the potassium ion, K+, has the same electronic configuration as the chloride ion, Cl-, and the same electronic configuration as an atom of argon, Ar. Therefore, Ar, Cl-, and K+ are said to be isoelectronic species.

For an atom or ion to be isoelectronic to another atom or ion, the SAME amount of electrons must be present. The Cl- anion has 18 electrons. S2- also has 18 electrons so it is isoelectronic. The K+ cation has also lost one electron from its original 19, making it isoelectronic with Cl- as well.

5. What gases was produced from decomposing hydrogen peroxide? What non-gaseous product formed from the reaction

Answers

Answer:

H2O

Explanation:

The equation for the decomposition of hydrogen peroxide is shown below;

2H2O2(l)-----> 2H2O(l) + O2(g)

Hence, the decomposition of H2O2 yields oxygen gas and water. Water is a non gaseous product of the reaction as clearly seen in the equation above.

(ii) What is the approximate chlorine-carbon-chlorine bond angle in C2Cl4?​

Answers

Answer:

O

C

C

l

 

 

120

The central carbon is  

s

p

2

hybridized........

Explanation:

And thus  

C

l

C

C

l

and  

O

C

C

l

 

are  

120

to a first approximation.

Why this value? We look to  

VSEPR theory

. There are 3 regions of electron density around the central carbon, and the most stable geometry is a trigonal plane. While there is a  

carbonyl

group, i.e. a  

C

=

O

bond, the  

π

bond is conceived to lie above and below this trigonal plane.

The  

carbonyl oxygen

is likewise conceived to be  

s

p

2

-hybridized

, however, here, there are 2 lone pairs on the oxygen centre.

The approximate chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.

The bond angle is defined as the angle between ant two bonds emanating from a common atom.

The compound C2Cl4 is tetrachloroethene. The carbon atoms are sp2 hybridized in this molecule.

Recall that the bond angle of an sp2 hybridized carbon atom is 120°. Therefore the chlorine-carbon-chlorine bond angle in C2Cl4 is 120°.

From the perspective of the VSEPR theory, the geometry of each carbon atom in C2Cl4 is trigonal planar which implies a bond angle of 120°.

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I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.

Answers

Answer:

175

Explanation:

How many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution?​

Answers

17.5 grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution.

What is Solution?

A solution is a homogenous mixture composed of two or more substances. In a solution, the components are evenly distributed throughout the mixture, resulting in a uniform appearance and properties.

The substance that is present in the largest amount is called the solvent, while the other substances present in lesser amounts are called solutes. When the solute dissolves in the solvent, the resulting mixture is called a solution.

To calculate the grams of NaCl needed to prepare a 35.0% salt solution, we can use the formula:

grams of NaCl = (percent salt / 100) x grams of solution

grams of NaCl = (35.0 / 100) x 50.0

grams of NaCl = 0.35 x 50.0

grams of NaCl = 17.5

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A teaspoon of salt, NaCl has a mass of about
5.0 g. How many formula units are in a
teaspoon of salt?

Answers

Answer: The answer is 5.15x10^22

Explanation:

The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.

Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).

Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.

Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.

The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.

To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:

Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:

Moles = Mass / Molar mass

Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]

Calculate the number of formula units using Avogadro's number:

Formula units = [tex]Moles \times Avogadro's number[/tex]

Let's perform the calculation:

Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]

Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]

Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.

Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].

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It took 70 seconds for 280cm³ of nitrogen to diffuse through a membrane. If Carbon(IV)Oxide is allowed to diffuse through the same membrane, how long will it take the gas to do so ? ​

Answers

Answer:

t = 125.3 seconds

Explanation:

Molar mass of CO2 = 12+2(16) = 66

Molar mass of N2 = 2(14)= 28

rate of diffusion of N2 = volume/ time = 280cm³/70s

= 4cm³/s

let rate of CO2 = rate of diffusion of CO2 = volume/time

= 400/t

Using Graham's law of diffusion,

rN2/rCO2 = √M(CO2)/M(N2)

4/400/t =√44/28 = 4t/400= √11/7

t/100 = 1.253 , t= (100)(1.253)

t = 125.3 seconds

hence it takes CO2 125.3 seconds to diffuse through the membrane

what are the strengths in the bonds of potassium bromide

Answers

Answer: Potassium Bromide (KBr) The Ionic bond formed between Potassium and Bromine is created through the transfer of electrons from Potassium (metal) to Bromine (nonmetal).

Explanation: this type of structure departs strongly from that expected for ionic bonding and ... whose roots go back to Max Planck's explanation in 1900 of the properties of ... types of interactions between elementary particles (the strong force, the weak force, ...

how many moles of h2 can be made from the complete reaction of 3.5 moles of al?
Given: 2Al+6HCL 2Alcl3+3h2

Answers

Answer:

From the given equation, we can see that for every 2 moles of Al, we get 3 moles of H2

So, we can say the the number of moles of H2 is 3/2 times the number of moles of Al

We are given the number of moles of Al and we have to find the number of moles of H2

We have deduced the relationship:

Moles of Al * 3 / 2 = Moles of H2

Replacing the variables with given values

3.5 * 3 / 2 = Moles of H2

Moles of H2 = 5.25 moles

Which is a chemical property of milk
A. Milk has a ph ranging from 6.4 to 6.8
B. Milk spoils when left unrefrigerated
C. Milk boils at about 212F
D. Milk curdles when mixed with vinegar

Answers

Answer:

C. Milk boils at about 212F

Explanation:

The principal constituents of milk are water, fat, proteins, lactose (milk sugar) and minerals (salts). Milk also contains trace amounts of other substances such as pigments, enzymes, vitamins, phospholipids (substances with fatlike properties), and gases.


C is the answer milk boils about 212F

To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?

Answers

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

How to calculate calories

Answers

Answer:If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2.

If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.

Explanation:

how many moles are in a 4.2 gram gold sample

Answers

The answer is 196.96655 hope this helps

If 0.4743 moles of H2O are produced, how many grams of VOCl3 will also be produced?
(V2O5 + 6 HCl → 2 VOCl3 + 3 H2O)

Answers

Answer:

i hope this helps. sorry if it totally doesn't

When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have:
a. different electron configuration from each other.
b. an unchanged electron configuration.
c. the same charge.
d. the same electron configuration (isoelectronic) as neon.
[Definition: The word isoelectronic means that when you write out the electron configuration they are the same. An exam would be He and Li whereby both of them have 2 electrons and therefore they are both are 1s2 in their electron configurations.]

Answers

Answer: d. the same electron configuration (isoelectronic) as neon.

Explanation:

Isoelectronic species are defined as the molecules which have the same number of electrons.

Atomic number of nitrogen is 7 and thus has 7 electrons. Nitrogen has electronic configuration of 2,5 and thus can gain 3 electrons and thus [tex]N^{3-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of oxygen is 8 and thus has 8 electrons. Oxygen has electronic configuration of 2,6 and thus can gain 2 electrons and thus [tex]O^{2-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of flourine is 9 and thus has 9 electrons. Flourine has electronic configuration of 2,7 and thus can gain 1 electron and thus [tex]F^{-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of sodium is 11 and thus has 11 electrons. Sodium has electronic configuration of 2,8,1 and thus can lose 1 electron and thus [tex]Na^{+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of magnesium is 12 and thus has 12 electrons. Magnesium has electronic configuration of 2,8,2 and thus can lose 2 electrons and thus [tex]Mg^{2+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

Atomic number of aluminium is 13 and thus has 13 electrons. Aluminium has electronic configuration of 2,8,3 and thus can lose 3 electrons and thus [tex]Al^{3+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)

the egents erosion and explain how each of them causes erosion​

Answers

Answer:

Wind, water, gravity, and ice

Explanation:

Water can erode soil material. Especially if the soil is bare, dry and erodible erosion via rain particles can occur. Wind is another factor that cause soil erosion. Dry soil particles( Especially if they are fine) can move to other areas if wind exists. Ice is another issue on erosion. Ice, when it is melting, can carry soil particles.

A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water weighing 15.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup.

Required:
Calculate the amount of salt dissolved in each liter of seawater.

Answers

Answer:

Amount of salt dissolved in each liter of seawater = 40 g

Explanation:

According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.

Also, the volume of seawater displaced equals the volume of freshwater in the cup.

From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.

Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup

Mass of freshwater = density of freshwater * volume

density of freshwater = 1 g/mL; volume = 375 mL

mass of freshwater = 375 mL * 1 g/mL = 375 g

mass of seawater = 375 + 15 = 390 g

mass of salt in 375 mL seawater = mass of seawater - mass of freshwater

mass of salt = (390 - 375) g = 15 g

Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g

Therefore, amount of salt dissolved in each liter of seawater = 40 g

Which is the product of that reaction

Answers

Answer:

B

Explanation:

Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.

Answers

Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.

What is an electric circuit?

The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.

In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.

In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.

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