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Fittoniya [83]
3 years ago
12

Trisha and her lab partner were conducting a variety of experiments to produce gases: hydrogen, oxygen, and carbon dioxide. In o

ne experiment, they added a piece of magnesium ribbon to 10 milliliters of hydrochloric acid. They observed bubbles being produced and did a variety of tests to identify the escaping gas; it proved to be hydrogen The reaction is represented by the following equation: Mg + 2HCl → MgCl2 + H2(g) Trisha and her lab partner filled a beaker with 150 milliliters of a 0.10 M HCl solution. How many moles of HCl are in the beaker? Remember, molarity = moles liter . A) 0.015 mol HCl B) 0.150 mol HCl C) 6.67 mol HCl D) 15 mol HCl
Chemistry
1 answer:
GaryK [48]3 years ago
7 0

<u>Given:</u>

Volume of HCl = 150 ml

Molarity of HCl = 0.10 M

<u>To determine:</u>

The # moles of HCl

<u>Explanation:</u>

The molarity of a solution is the number of moles of a solute dissolved in a given volume

In this case:

Molarity of HCl = moles of HCl/volume of the solution

moles of HCl = Molarity * volume = 0.10 moles.L-1 * 0.150 L = 0.015 moles

Ans: A)

Moles of HCl is 0.015

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Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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3 years ago
As temperature increases, the amount of solute that a solvent can dissolve increases. True False
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<span>As temperature increases, the amount of solute that a solvent can dissolve increases.</span>



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At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091

<u>Initial:</u>         0.225      0.225

<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

K_c=\frac{[ICl]^2}{[Cl_2][I_2]}

Putting values in above equation, we get:

9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0
3 years ago
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