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Olegator [25]
4 years ago
5

Convert 0.320 atm to mmHg.

Chemistry
2 answers:
Tatiana [17]4 years ago
7 0

Answer:  243.2 milometer of mercury

Explanation:

earnstyle [38]4 years ago
4 0

Answer:

243.2

Explanation:

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What is the h+ of a solution with a ph of 5.6
7nadin3 [17]
Just have to do antilog

[H+]= 10^-5.6
6 0
3 years ago
When is di- used in the name of a hydrocarbon?
harkovskaia [24]
Di- is used when you are naming organic compounds. If you have the same substituent repeated twice in the compund
For example: CH3-CH(CH3)-CH2-CH(CH3)-CH3
This will be named 2,4-dimethylpentane
6 0
3 years ago
The vaporization of Br2 from the liquid gas to the solid state requires 7.4 kcal/nil what is the triangleH for this process
SSSSS [86.1K]

The initial state of the compound Br2 is liquid. The final state is then solid. We know for certain that the liquid state has higher energy as compared to solid state. Due to this fact, Br2 must release energy to be in solid form. Therefore, the change in enthalpy is:

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5 0
3 years ago
PLEASE HELP :-)
timofeeve [1]
We can solve this problem using the long hand solution, wherein we 1 by 1 analyze the different equilibrium reactions or by simply using the Henderson Hasselbach equation. The equation is 

pH = -log(pKa) + log (salt/acid)

since the acid and the salt have the same concentration, the log (salt/acid) term is equal to zero. 

thus 

pH = -log(1.73*10-5)
pH = 4.76

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8 0
4 years ago
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

7 0
3 years ago
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