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natka813 [3]
2 years ago
10

Question 1 of 7

Chemistry
2 answers:
Nata [24]2 years ago
8 0

Answer:

b

Explanation:

nlexa [21]2 years ago
3 0

Answer:

Explanation:B ice melting is a phiyscal change and the mass of the substance remains the same

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1. Addition of which of the following will increase the solubility of CaCO3 in water? Consider the equilibrium process:
kolbaska11 [484]

Answer: HCl

Explanation:

calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:

CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)

This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.

6 0
3 years ago
What mass of AI2O3 forms from 16 g O2 and excess AI?
vlada-n [284]

Answer:

2Al+1.5O2→Al2O3

Thus, 2 mol of Al combine with 1.5 mol of oxygen to form 1 mol of Al2O3.

2 mol of Al corresponds to 2×27=54 g.

Thus, the weight of Al used in the reaction is 54 g.

4 0
2 years ago
Although fusion is an energetically favorable reaction for light nuclei, it does not occur under standard conditions on Earth be
hammer [34]
The best and most correct answer among the choices provided by your question is the first choice.

<span>Although fusion is an energetically favorable reaction for light nuclei, it does not occur under standard conditions on Earth because: fission reactions are always favored over fusion reactions.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
3 years ago
Read 2 more answers
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
Using the 5M NaCI and 10% glucose stock solutions you generated, how would you prepare 100 ml of a solution that is both 150mM N
fredd [130]

Answer:

Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

Explanation:

  • In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:

0.1 L * 0.150 M = 0.015 moles of NaCl

With those moles we can calculated the required volume, using the concentration of the stock solution:

0.015 mol / 5 M = 0.003 L = 3 mL.

  • To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.

So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

4 0
3 years ago
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