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Volgvan
4 years ago
12

Balance the following equation by inspection: Cu + HNO3 --> Cu(NO3)2 + H2O +NO

Chemistry
2 answers:
lorasvet [3.4K]4 years ago
8 0

Answer:

3Cu + 8HNO_{3} → 3Cu(NO_{3} )_{2} +4H_{2}O + 2NO

Explanation:

Considering the number of each atom in both reactants, it must be equal to the number of each atom in the product. Thus, the number of copper atom, hydrogen atom, nitrogen atom and oxygen atom must be equal on both sides of the equation.

pashok25 [27]4 years ago
7 0

Answer:

f is this

Explanation:

You might be interested in
glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate AT
ddd [48]

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula:  ∆Go ’= -RTln K’eq

where,  

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

=  - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

=  - 8.315 x 298 x 6.79

=  - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

=  -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

6 0
3 years ago
How much energy is required to raise the temperature of a 300.0
Montano1993 [528]

Answer:

Q = 1455.12 Joules.

Explanation:

Given the following data;

Mass = 300 grams

Initial temperature = 22.3

Final temperature = 59.9°C

Specific heat capacity = 0.129 J/gºC.

To find the quantity of energy;

Q = mcdt

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt represents the change in temperature.

dt = T2 - T1

dt = 59.9 - 22.3

dt = 37.6°C

Substituting the values into the equation, we have;

Q = 300*0.129*37.6

Q = 1455.12 Joules.

8 0
3 years ago
The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38
adell [148]

Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = 1.21x10^{-3} M

7 0
3 years ago
Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074
Hoochie [10]

Answer:

0.3793 M

Explanation:

The unknown metal is zinc. So the equation of the reaction is;

Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)

From Nernst equation;

E = E° - 0.0592/n log Q

[Cu2+] = 0.050179 M

n = 2

[Zn^2+] = ?

E = 1.074 V

E° = 0.34 - (-0.76) = 1.1 V

Substituting values;

1.074  = 1.1 - 0.0592/2 log [Zn^2+]/0.050179

1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179

-0.026 = -0.0296  log [Zn^2+]/0.050179

-0.026/-0.0296 = log [Zn^2+]/0.050179

0.8784 =log [Zn^2+]/0.050179

Antilog(0.8784) = [Zn^2+]/0.050179

7.558 =  [Zn^2+]/0.050179

[Zn^2+] = 7.558 * 0.050179

[Zn^2+] = 0.3793 M

4 0
3 years ago
A chemistry graduate student is given 500. mL of a 0.20 M chloroacetic acid (HCH2ClCO2) solution. Chloroacetic acid is a weak ac
zhenek [66]

Answer:

12 g of choloracetic acid

Explanation:

The buffer equilibrium is:

HCH₂ClCO₂ ⇄ CH₂ClCO₂⁻ + H⁺

pka= -log ka =

Ka: 1,3x10⁻³ = [CH₂ClCO₂⁻] [H⁺] / [HCH₂ClCO₂]

By Henderson-Hasselbalch equation:

pH = pka + log₁₀ [A⁻] / [HA]

3,01 = 2,89 + log₁₀ [A⁻] / [HA]

1,318 = [A⁻] / [HA]

As molar concentration of chloroacetic acid (HA) is 0,20M

[A⁻] = 0,26 mol/L

The volume is 500 mL ≡ 0,5 L

0,26mol/L × 0,5 L = 0,13 moles of chloroacetic acid. In grams:

0,13 mol × (94,5g / 1mol) = <em>12 g of choloracetic acid</em>

<em></em>

I hope it helps!

3 0
3 years ago
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