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RSB [31]
3 years ago
11

7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?​

Chemistry
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

The volume is

<h2>180 mL</h2>

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

<h3>P _{1}  V _{1} = P _{2}V _{2}</h3>

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

<h3>V _{2}  = \frac{P _{1}  V _{1}}{P _{2}  }</h3>

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3>V _{2} =  \frac{300 \times 300}{500}  \\  =  \frac{90000}{500}  \\  =  \frac{900}{5}</h3>

We have the final answer as

<h3>180 mL</h3>

Hope this helps you

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a technician mixes 80 ml of a 5% solution with 10 ml of water. what is the final percentage strength of the solution prepared?
Vinvika [58]

A technician mixes 80 ml of a 5% solution with 10 ml of water.  the final percentage strength of the solution prepared is 40 %.

given that :

8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL

therefore, 80 × 5 = 10 × x %

x % = 40 %

Therefore, the final percentage strength of the solution is 40 %

Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water.  the final percentage strength of the solution prepared is 40 %.

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5 0
1 year ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
2 years ago
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tankabanditka [31]
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Plugging in the given values:
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(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ 
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