2.
(3x – 4y = -10
(6x + 3y = –42
SOLUTION:

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.

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.

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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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.

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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Answer:

Mass of nitrogen dioxide produced = 4.6 g

Explanation:

Given data:

Volume of ammonia = 2.30 L

Mass of nitrogen dioxide produced = ?

Solution:

Chemical equation:

4NH₃ + 7O₂     →      4NO₂ + 6H₂O

Number of moles of ammonia at STP:

PV = nRT

n = PV/RT

n = 1 atm × 2.30 L / 0.0821 atm.L/K.mol × 273 K

n = 2.30 atm .L / 22.414 atm.L/mol

n = 0.1 mol

Now we will compare the moles of ammonia with nitrogen dioxide from balance chemical equation.

                NH₃            :             NO₂

                 4                :               4

                 0.1             :              0.1

Mass of NO₂:

Mass = number of moles  × molar mass

Mass = 0.1 mol  × 46 g/mol

Mass = 4.6 g

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

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!

Answer:

Formation of covalent bond structures. The image is essentially a Lewis dot structure.

The image of HOH has depicted the covalent bond formation in water. Thus, option B is correct.

The image has been the Lewis structure of the water molecule. It has been consisted of two hydrogen atoms bonded to oxygen molecules.

The oxygen has a presence of 4 dots that have been the representation of the valence electrons.  Thus, oxygen has been consisted of 2 lone pairs.

The water molecule has the presence of shared electrons between hydrogen and oxygen. The bond formed by the sharing of electrons has been covalent.

Thus, the image of HOH has been the representation of the formation of covalent bonds in water. Thus, option B is correct.

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Answer:

Mercury Atom

Explanation:

If you are talking about what has 121 neutrons and 80 electrons, then it would be a Mercury Atom. Hope this helps!

Answer:

[tex]m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

[tex]n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2[/tex]

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

[tex]m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]

Best regards.

The mass of  Cd(OH)₂ that would be formed is 36.6 g

From the question,

We are to determine the mass of Cd(OH)₂ that would be formed

The given balanced chemical equation for the reaction is  

2NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2NaNO₃(aq)

This means

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂ and 2 moles of NaNO₃

Now, we will determine the number of moles of each reactant present

Mass = 20.0 g

Molar mass = 39.997 g/mol

From the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Number of moles of NaOH present = [tex]\frac{20.0}{39.997 }[/tex]

Number of moles of NaOH present = 0.50 mole

Volume = 0.750 L

Concentration = 1.00 M

From the formula

Number of moles = Concentration × Volume

∴ Number of moles of Cd(NO₃)₂ present = 1.00 × 0.750

Number of moles of Cd(NO₃)₂ present = 0.750 mole

Since,

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂

Then,

0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂

∴ The number of moles of Cd(NO₃)₂ that reacted is 0.25 mole

Now,

From the balanced chemical reaction

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂

Then,

0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂ to produce 0.25 mole of Cd(OH)₂

∴ Number of moles of Cd(OH)₂ formed is 0.25 mole

Now, for the mass of Cd(OH)₂ that would be formed

From the formula

Mass = Number of moles × Molar mass

Molar mass of Cd(OH)₂ = 146.43 g/mol

∴Mass of Cd(OH)₂ formed = 0.25 × 146.43

Mass of Cd(OH)₂ formed = 36.6075 g

Mass of Cd(OH)₂ formed≅ 36.6 g

Hence, the mass of  Cd(OH)₂ that would be formed is 36.6 g

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Answer:

B

Explanation:

Answer:

mass = density × volume = 0.67 × 1.00 = 0.67 kg = 670 g. The molar mass of octane, C8H18, is 8 × 12.01 (C) + 18 × 1.008 (H) = 114.224.

Explanation:

Answer:

Solution-

From the question the volume of water = 18 L = 18000 mL

Now we can find the mass of water = (volume of water) * (density of water)

mass of water = (18000 mL) * (1.00 g/mL)

mass of water = 18000 g

Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)

putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)

heat required to boil water = 5821617.6 J

heat required to boil water = 5821.62 kJ

The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)

Heat given by combustion = (5821.62 kJ) / (19.4 /100)

Therefore the heat given by combustion = (5821.62 kJ) / (0.194)

Heat given by combustion = 30008.35 kJ

As we know that the enthalpy of combustion of methane = 802.5 kJ/mol

The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)

moles methane used = (30008.35 kJ) / (802.5 kJ/mol)

So the moles methane used = 37.39 mol

Now the mass methane = (moles methane used) * (molar mass methane)

The mass methane = (37.39 mol) * (16.04 g/mol)

The mass methane = 599.74 g

Now the volume methane = (mass methane) / (density of methane)

volume methane = (599.7356 g) / (0.660 g/L)

volume methane = 908.69 L

hope helped!!

plz mark brainliest:DD

The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L

Using the given data :

Volume of water = 16.5 L = 16500 mL

mass of water = 16500 g  ( 16500 mL * 1.00 g/mL )

Density of methane = 0.668 g/L

Density of water = 1.00 g/mL

First step : determine the heat needed to boil the water

Heat required = mass of water * specific heat water * ( Δ T )

                        = 16500 * 4.184 * ( 100 - 20.4 ) =  5495265.6 J

                        = 5495.265 kJ

∴ Heat required to boil water = 5495.265 kJ

next step ; determine the heat given by combustion

heat given by combustion = ( Heat required to boil water) / ( % of heat generated )

Heat given by combustion = ( 5495.265 ) / ( 17.9 % )

                                             = 30699.80 kJ

Enthalpy of methane combustion = 802.5 kJ/mol

∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol

next ; determine the mass of methane ( natural gas )

= ( moles of methane used ) * ( molar mass )

= 38.26 * 16.04 g/mol = 613.69 g

Final step : Calculate the volume of natural gas is needed to boil the water

= mass of natural gas / density of methane

= 613.69 g / 0.668 g/L

= 918.70 L

Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is  918.70 L .

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Answer:

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

Explanation:

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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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

Answer:

A

Explanation:

As the space in a container gets, smaller the volume decreases because volume is a measure of the space that the particles can occupies inside.

Volume is the space occupied by a substance and which is directly proportional to temperature according to Boyles law and inversely proportional to the pressure.

As the temperature increases, volume increases whereas the pressure increases, volume get decreased.  As there are more particles but the space inside a container is small, then their volume is said to be smaller.

Hence, option C is the answer.

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Answer:

175

Explanation:

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

Answer:

48000

Explanation:

maths

Answer:

48000

Explanation:

(5x10^3) = 5 × 1000 = 5000

(4.3x10^4) = 4.3 × 10000 = 43000

5000 + 43000 = 48000

Answer:

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

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.

Answer:

Which one of the table is element X?

Explanation:

I have no idea and help!!!!!!!!!!!!!!!!!!!!!!!!!

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, ...

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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Answer:

6 protons 8 neutrons 5 electrons

Explanation:

protons are blue neutrons are red electrons are green

Answer:

0.75

Explanation:

3/4=0.75 so therefore its 0.75

Answer:

Explanation:

moles of potassium dichromate = .0250 x .026 = 65 x 10⁻⁵ moles

1 mole of potassium dichromate reacts with 6 moles of Fe⁺²

65 x 10⁻⁵ moles of potassium dichromate will react with

6 x 65 x 10⁻⁵ moles of Fe⁺²

= 390 x 10⁻⁵ moles

390 x 10⁻⁵ moles are contained in 25 mL of solution

molarity of solution = 390 x 10⁻⁵ / 25 x 10⁻³

= 15.6 x 10⁻² M  .

Answer:

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

Hope this helps you

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.)

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:

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.