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3 years ago

Write the balanced chemical equations of each reaction:

Chemistry

1 answer:

Darya [45]3 years ago

3 0

Answer:

A. HBr + KHCO3 >>> H20 + CO2 + KBr

B. 3SnO + F3N >>> 3F2Sn + N2O3

C. CH4 + O2 >>> CO2 + H2O

D. AlPO4 + 3LiNO3 >>> Al(NO3)3 + Li3PO4

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What is the mass of 23450 L of hydrogen gas at STP

WINSTONCH [101]

Remember that in this case pressure is equal to 1.00 atm and temperature is equal to 273.15K. So,
P
V
=
n
R
T
→
n
=
P
V
R
T
=
1.00
a
t
m
⋅
7.0
L
0.082
a
t
m
⋅
L
m
o
l
⋅
K
⋅
273.15
K
=
0.31
Since we know hydrogen's molar mass (
2.0
g
m
o
l
), we can determine the mass
m
H
2
=
n
⋅
m
o
l
a
r
.
m
a
s
s
=
0.31
m
o
l
e
s
⋅
2.0
g
m
o
l
=
0.62

g
If indeed you are dealing with STP, remember that, under these conditions, 1 mole of any ideal gas occupies
22.4
L
. So,
n
=
V
V
m
o
l
a
r
=
7.0
L
22.4
L
=
0.31
moles
And, once again,
m
=
0.31
⋅
2.0
=
0.6

4 0

3 years ago

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ

Ahat [919]

Answer: The Gibbs free energy of the reaction is 21.32 kJ/mol

Explanation:

The chemical equation follows:

$\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}$

$\text{[Oxaloacetate]}=0.130mM$

$[NADH]=2.0\times 10^2mM$

$\text{[Malate]}=1.37mM$

$[NAD^+]=490mM$

Putting values in above expression, we get:

$\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol$

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0

3 years ago

Please help!!! Its not chemistry is science.

laiz [17]

You would want to know everything just Incase anything happens because maybe you get lost

7 0

4 years ago

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima

svetlana [45]

A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$ = 0.10 M

The chemical equation for this reaction is :

$\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = $\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}$ = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

$\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}$

Initial 0.10 0.035 0.10 -

Change -0.035 -0.035 + 0.035 -

Equilibrium 0.065 0 0.135 -

By using Henderson-Hasselbalch equation:

The pH of this solution = pKa + log $\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}$

The pH of this solution = 4.74 + log $\mathtt{\dfrac{0.135}{0.065}}$

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06 to two decimal places

7 0

3 years ago

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iogann1982 [59]

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