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Advocard [28]
3 years ago
14

20 pts. Hydrogen and oxygen react to form water.

Chemistry
1 answer:
FromTheMoon [43]3 years ago
3 0
A. would be H20 because O is 2- and H is 1+ so it needs to balance to 0.
sorry I don't know the others but I hope this helps with letter A
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A balanced chemical equation used to prepare ammonium carbonate, (nh4)2co3 , is: 2 n h 3 ( g ) + c o 2 ( g ) + h 2 o ( l ) ⟶ ( n
Tems11 [23]

Answer:

This a simple stoichiometry problem using the ideal gas law.  

First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol  

Now use a molar conversion using the balanced equation,  

1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas  

PV=nRT  

V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L

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3 years ago
How many elements have been discovered so far by scientists ​
Inessa [10]

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118 elements have been discovered so far

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3 years ago
List 3 scenarios for a substance being classified as a weak acid
malfutka [58]
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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

8 0
3 years ago
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