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3 years ago

A mole of hydrogen atoms has 6.02 x 10^23 atoms. It occupies 22.414 L. How many hydrogen atoms are in 50.00 mL?

Chemistry

1 answer:

Greeley [361]3 years ago

3 0

Answer:

This is a simple case of ratios. (1 mol)/(22.4 L)=(n mol)/(.025 L) Then we cross multiply and we get 22.4n=.025 We divide each side by 22.4 to find n=.001116 mol Then to convert the moles to atoms we multiply, and cross-cancel the units (.001116 mol)/1 xx (6.02 xx 10^23 atms)/(1 mol) and we have 6.72 xx 10^20 atoms. I've found the trick of cross-cancelling units to be a very effective mnemonic, it always makes sure you carry out the correct calculation to find the desired units.

Explanation:

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What is Delta.Gsystem for the system that is described by the following data? Delta.Hsystem = –345 kJ, T = 293 K, Delta.Ssystem

ryzh [129]

Answer:

-209 kJ

Explanation:

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6 0

4 years ago

What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as

serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

CaCl2 ---> Ca2+ + 2 Cl-

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

Cl- = 0.4 moles (ANSWER)

6 0

3 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

3 years ago

Which of the following statements are true of acids and bases?

gtnhenbr [62]

A. Acids increase the number of OH- ions in a solution. FALSE
B. Bases increase the number of OH- ions in a solution. TRUE
C. Acids increase the number of H+ ions in a solution. TRUE
D. Bases increase the number of H+ ions in a solution. FALSE

8 0

3 years ago

How can you measure the volume of air inside of a balloon?

Pani-rosa [81]

You could let the air out of the balloon while it is under the water with a container filled with water upside down over it. And measure the water displacement.

8 0

3 years ago