A 12.1 M HCl solution with a density of 1.19 g/mL has a molality of 16.1 m.
What is molality?Molality is defined as the total moles of a solute contained in a kilogram of a solvent.
Let's suppose we have 1 L of the solution.
Step 1: Calculate the moles and mass of HCl (solute) in 1 L of solution.The solution is 12.1 M, that is, there are 12.1 moles of HCl in 1 L of solution.
The molar mass of HCl is 36.465 g/mol.
12.1 mol × 36.465 g/mol = 441 g
Step 2: Calculate the mass corresponding to 1 L of solution.The density of the solution is 1.19 g/mL.
1000 mL × 1.19 g/mL = 1190 g
Step 3: Calculate the mass of water (solvent) in 1190 g of solution.In 1190 g of solution, there are 441 g of HCl.
mWater = 1190 g - 441 g = 749 g = 0.749 kg
Step 4: Calculate the molality (b) of the solution.b = 12.1 mol / 0.749 kg = 16.1 m
A 12.1 M HCl solution with a density of 1.19 g/mL has a molality of 16.1 m.
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Why NH2 is a bad leaving Group?
Explanation:
Consider that strong bases are strong nucleophiles, and thus unstable with the negative charge associated with dissociation from the molecule. Hence, strong bases are generally very poor leaving groups, but can be converted into good leaving groups that are more stable with the associated negative charge.
A car engine has a displacement of 536 cubic inches (in^3). What is the engines displacement in nm^3? Use dimensional analysis and show all work
The engine displacement in cubic nanometer (nm³) is 8.785×10²⁴ nm³
Data obtained from the question Displacement in in³ = 536 in³Displacement in nm³ =? Conversion scale1 in³ = 1.639×10²² nm³
With the above convesion scale we can obtain the displacement in nm³
How to determine the displacement in nm³1 in³ = 1.639×10²² nm³
Therefore,
536 in³ = 536 × 1.639×10²²
536 in³ = 8.785×10²⁴ nm³
Thus, the displacement in nm³ is 8.785×10²⁴ nm³
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Need help with chemistry problem
Answer:
b option sir sidd dichhi good book goo
The Lewis structure below represents the compound ___________________.
Group of answer choices
dioxygen carbide
carbon tetraoxide
carbon dioxide
carbon monoxide
Answer:
Carbon Dioxide
Explanation:
This is a covalent compound so to name it you have one carbon, no prefix needed, & 2 oxygen the prefix for 2 is Di-.
Giving Braninliest!
Which of the following circuits is built correctly?
Answer:
number 2 becaus 1 and 3 look simler.
Explanation:
state four temperature scales
Answer: Celsius, Fahrenheit, Kelvin, and Rankine.
Hope this helped you!
Compared to protons, electrons have:
a. much smaller mass and oppisite charge.
b. about the same mass and opposite charge.
c. much larger mass and the same charge.
d. much larger mass and oppisite charge.
Explanation:
a. much smaller mass and oppisite charge.
Which of the following is an example of an ionic compound?
A.) Carbon Dioxide
B.) Oxygen
C.) Sodium Chloride
D.) Water
Answer:
C.) Sodium Chloride
Explanation:
Table salt is an example of an ionic compound. Sodium and chlorine ions come together to form sodium chloride or NaCl.
What is the difference between salinity and chlorinity?
Answer:
salinity is the concentrations of salts in water or soils, chlorinity is a measure of the concentration of halides in one kilogram of seawater. Explanation:
srry if its wrong ;-;
Answer:
Chlorinity and salinity are both measures of the saltiness of sea water. The relationship can be expressed mathematically, as salinity is equivalent to 1.80655 times the chlorinity.
Conclusion:
They both are the measures for the same thing, except salinity is equivalent to 1.80655 times the chlorinity.Learn more:
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Mathematics
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Answered by: ms115
vanadium hydroxide ionic or covalent?
Answer:
Vanadium in given compound is metal and oxygen is non-metal so, the given compound is a binary ionic compound.
How do you separate and collect salt and water after a neutralisation reaction?
Explanation:
water can be separated from salt solution by simple distillation. This method works because water has a much lower boiling point than salt. When the solution is heated, the water evaporates. It is then cooled and condensed into a separate container.
How many grams of magnesium bromide (MgBry) are in 5.38 x 1024 formula units
of the substance?
Answer:
is 5509.12
Explanation:
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Calculate the number of atoms/molecules present in
(a) 1 mole of sulphur.
(b) 2 moles of oxygen gas.
Answer:
1 mole contains 6.02 ×10^23 atoms
1 mole of Sulphur will contain 6.02 ×10^23 atoms
2moles of. Oxygen will contain 2 x 6.02×10^23 atoms. This gives 12.04 × 10^23 atoms
What is the empirical formula of C6H12O6?
C2H4O2
CH2O
CH2O2
C2H4O
Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O
Explanation:
Part B
The image models what happened after each cup was placed over an effervescent tablet. Note the changes in the water level and the air space in both glasses. The tablets are shown to help see how the experiment was setup, but the model represents the air bubbles after the tablets fully dissolved. Write down your observations in the space provided.
Answer:
The Changes I see is the Hot glass have less air bubbles than the Cold Glass. The Cold Glass has more air bubbles than the Hot Glass.
I hope this helps. I tried my best!! Good Luck!!
Find the percent composition of water in barium chloride dihydrate. Round your answer to the
nearest whole number.
A)17%
B)8%
C)85.3%
The percent composition of water will be 15%
Percent compositionThe percent composition of a component of a compound is found by dividing the mass of the component by the mass of the compound.
In other words:
Percent composition = mass of component/mass of substance x 100
In this case, the formula for barium chloride dihydrate is [tex]BaCl_2.H_4O_2[/tex]
The molar mass of the compound is 244.26 g/mol
The molar mass of the water component is 18 x 2 = 36 g/mol
Thus, the percent composition of water in the compound can be found, such that:
Percent composition of water = 36/244.26 x 100
= 15% to the nearest whole number.
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9. Select 2 that apply.
What statements do NOT apply to Hess's Law ? Select all that apply.
To find the enthalpy of formation of a compound, you can use the enthalpy changes of reactions that involve the elements that make up the
compound, even if those reactions are not necessarily intermediate steps in the production of that compound
Known enthalpy changes can be combined to calculate unknown enthalpies of reaction
If a reaction contains intermediate steps, the overall enthalpy change is equal to the last step where the enthalpy changed.
Thermochemical equations may be reversed, if necessary, to calculate unknown enthalpies, as long as the sign of the enthalpy value also changes.
No thermochemical equation can ever be reversed to calculate enthalpy, even if the value is changed.
If a reaction contains intermediate steps, the overall enthalpy change of the reaction is equal to the sum of the enthalpy changes of the intermediate
steps.
The statements that are applicable to Hess's Law include:
If a reaction contains intermediate steps, the overall enthalpy change is equal to the last step where the enthalpy changed.No thermochemical equation can ever be reversed to calculate enthalpy, even if the value is changed.What is Hess's Law?Hess's Law is also referred to as Hess's law of constant heat summation and it was named after a Swiss-born Russian chemist called Germain Hess.
Hess's Law states that the energy change experienced in an overall chemical reaction is equal to the sum of the energy changes in each chemical reactions that it contains.
In conclusion, the enthalpy change of a chemical reaction which is the heat of reaction at constant pressure is independent of the pathway between its initial and final states.
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How can we explain the motion of objects?
Someone can help me please??
Answer:
first option - liquid particles gain kinetic energy and start moving faster around each other
Explanation:
this is a rising section of the graph, meaning that the state of matter is not changing but it warming up and increasing particle movement. This increase in movement is explained by the increase in kinetic energy. (when the kinetic energy rises, the movement rises)
The difference between the amount of heat actually released upon the hydrogenation of benzene and that calculated for the hydrogenation of an imaginary cyclohexatriene is called the _________ of benzene.
From what we know, we can confirm that the difference measured by the experiment in the heat released when compared to the hydrogenation of an imaginary cyclohexatriene is called the stabilization energy of benzene.
What is stabilization energy?This is a term used to describe the difference in measured energy of a chemical structure compared to the theoretical energy amounts expected to be measured. In this experiment, the heat that was released did not match the calculated energy expected, which is a prime example of stabilization energy at work.
Therefore, we can confirm that the difference measured by the experiment is called the stabilization energy of benzene.
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how many mL are in 0.0543L
Answer:
Multiply 0.0543 L by a 1000 to get to mL, move the decimal point 3 times to the right, and you get 54.3 mL
Write the pressure equilibrium constant expression for the reaction N2(g)+3H2(g)=2NH3(g)
Answer:
[tex]keq=\frac{[NH_{3} ]^2}{[N_{2} ][H_{2} ]^3}[/tex]
Explanation:
how many grams of k2co3 are needed to make 255 ml of a 88.5 m solution
Answer: You need approximately 70 g of potassium carbonate.
Explanation:
Hope this helps :) Good Luck!
Pls help me before my mother takes away my phone
Answer:
sorry man i cant help you i tried my best im so so so sorry
Explanation:
Emi wants to see if she can melt a marshmallow in the sun. She plans to put the marshmallow in a container that will cause it to heat up quickly. In one or two sentences, describe what type of container and materials Emi should use.
I GIVE BRAINLY IF ANSWEREDD CORRECT
Answer:
emi should put it in a metal container and place directly to the sun emi should place two mirrors opposite it and that will increase the heat which will melt the Marsh mallow
A student ran the following reaction in the laboratory at 546 K:
COCl2 (g) CO (g) + Cl2 (g)
When she introduced 0.723 moles of COCl2 (g) into a 1.00 liter container, she found the equilibrium concentration of Cl2 (g) to be 3.38×10-2 M.
Calculate the equilibrium constant, Kc, she obtained for this reaction.
Kc =
Answer: For the reaction 2HCl(g) ==> H2(g) + Cl2(g), the Keq = [H2(g)][Cl2(g)]/[HCl(g)]^2
Explanation:
56.56
Two chemicals react in an endothermic reaction. What can be known about this reaction?
O The new substance will need more energy to form its chemical bonds than the old substance will release.
O More energy will be released from the old substance than the new substance will need to form its chemical bonds.
O A precipitate will form as a result of the reaction.
O A gas will form as a result of the reaction.
Answer:
it's a I took the test
Explanation
Answer:
A) The new substance will need more energy to form its chemical bonds than the old substance will release.
Explanation:
What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol
Answer:
58.92 g EDTA
Explanation:
315.1 mL = .3151 L
M = Moles / Liter
.3151 L x 0.5 mol EDTA x 374 g EDTA = 58.92 g EDTA
1 L EDTA 1 mol EDTA
The amount in grams of EDTA needed to make 315.1 mL of a 0.05 M is equal to 5.89 g.
Molar concentration
In a 0.05M solution, there are 0.05 moles of the substance per 1 liter, that is, in 315.1 ml we have:
[tex]\frac{0.05 mol}{xmol}= \frac{1000ml}{315.1ml}[/tex]
[tex]x = 0.016 mol[/tex]
Thus, 0.016 mol EDTA of is needed to prepare the solution.
Mass calculationTo calculate the mass in an amount of moles, we use the molar mass, so that:
[tex]m = MM \times mol[/tex]
[tex]m = 374 \times 0.016[/tex]
[tex]m = 5.89 g[/tex]
So, the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M is equal to 5.89 g.
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I need help with this please????
[tex]\tt \rho=\dfrac{m}{v}=\dfrac{7\times 10^2~g}{1.3~g/L}=538~L[/tex]
Someone can help me please??
Answer:
c-d
Explanation:
this is a section of the graph sloping down, meaning the state of matter isn't changing but is getting ready too (the actually changing of states happens in the flat sections). A-B is the gas cooling, b-c is the gas changing into liquid, which leads to the liquid cooling in the next section