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Leya [2.2K]
3 years ago
9

Use the conversion factor 1 amu = 1.66054× 10–24 g to answer the following questions. a) 1.67 x 10-24 g of neurons is how many (

amu)
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
7 0
You can use the mass of neuron divided by the mass of conversion factor: 1.67*10^(-25)/(1.66054*10^(-24))≈1 amu. So the answer is 1 amu. 
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Answer:

the answer is the first one (a)

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2 years ago
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A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the con
inn [45]

Answer:

The equilibrium concentration of I₂ is 0.400 M  and HI is 1.20 M, the Keq will be 0.112.

Explanation:

Based on the given information, the equilibrium reaction will be,  

2HI (g) ⇔ H₂ (g) + I₂ (g)

It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,  

= 4.00 mol/2.00 L

= 2.00 mol/L

Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.  

It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.  

Now the equilibrium concentration of HI will be,  

= 2.00 -2x  

= 2.00 - 2 × 0.400

= 1.20 M

The equilibrium concentration of I₂ will be,  

I₂ = x  

= 0.400 M

The equilibrium constant (Keq) will be,  

Keq = [H₂] [I₂] / [HI]²

= (0.400) (0.400) / (1.20)²

= 0.112

Thus, the Keq of the reaction will be 0.112.  

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3 years ago
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2 years ago
Hydrogen sulfide gas is combusted with oxygen gas to produce sulfur dioxide gas and water vapour. If there is 48.4 L of oxygen a
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Answer:

2H2S + 3O2 → 2SO2 + 2H2O

V(O2) = 48.4 L

p = 105 kPa = 1.036 atm

T = 190 + 273 = 463 K

Ideal gas law:

pV = nRT

n = \frac{pV}{RT}n=  

RT

pV

​

 

R = 0.08206 L×atm/mol×K

n(O2) = \frac{1.036 \times 48.4}{0.08206 \times 463}=1.319 \; mol=  

0.08206×463

1.036×48.4

​

=1.319mol

According to the reaction:

n(H2S) = \frac{2}{3}  

3

2

​   n(O2) = \frac{2}{3} \times 1.319 = 0.8798 \;mol  

3

2

​ ×1.319=0.8798mol

V = \frac{nRT}{p} \\ V(H_2S) = \frac{0.8798 \times 0.08206 \times 463}{1.036}=32.26 \;LV=  

p

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​ V(H  

2

​  S)=  

1.036

0.8798×0.08206×463

​  =32.26L

Answer: 32.26 L

Explanation:

5 0
3 years ago
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