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kvv77 [185]
3 years ago
9

Please help 10 POINTS!!!

Chemistry
1 answer:
salantis [7]3 years ago
5 0
The answer would be C as in cat and D as in dog your welcome i hope this help
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Small quantities of hydrogen can be prepared by the addition of hydrochloric acid to zinc. a sample of 195 ml of hydrogen was co
Vikentia [17]
Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:

PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂

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Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
5 0
3 years ago
Read 2 more answers
Convert this temperature from F to C<br><br> 26.6°c<br> 93.600<br> 62.2°C<br> 5.7°C
eduard
The answer for this is 26.6°c
3 0
3 years ago
20.00 mL of a 0.077 M solution of silver nitrate, AgNO3
melomori [17]
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8 0
3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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