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Amanda [17]
2 years ago
12

Calculate the pH if the [OH-] concentration is 5.9x10-1 M.

Chemistry
1 answer:
ANEK [815]2 years ago
7 0
Ph—1.763
Oh-58
Answers
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What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95
melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

Q = c \times m \times \Delta T

where,

  • c: specific heat capacity
  • m: mass
  • ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol

8 0
2 years ago
I need the answer fast plzzz!! NO LINKS
Arturiano [62]

Answer:

your answer is 12 hope it's correct answer

4 0
2 years ago
Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:
posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

Equilibrium concentration of NO = \frac{16.6g}{30g/mol\times 9.5L}=0.058M

Equilibrium concentration of NO_2 = \frac{22.5g}{46g/mol\times 9.5L}=0.051M

Equilibrium concentration of H_2O = \frac{189.0g}{18g/mol\times 9.5L}=1.10M

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

For the given chemical reaction:

2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)

The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

5 0
2 years ago
Which of the following scientists made an important contribution to quantum mechanics: (i) Bohr (ii) deBroglie (iii) Heisenberg
grandymaker [24]

Answer:

(i) Bohr; (ii) de Broglie; (iii) Heisenberg (v) Schrödinger

Explanation:

(i) Niels Bohr — 1913 — proposed that electrons travel in fixed orbits with <em>quantized energy levels</em> and that they jump from one energy level to another by absorbing or emitting quanta of light.

(ii) <em>Louis de Broglie</em> — 1924 — proposed the wave nature of electrons and suggested that all matter behaves as both waves and particles (<em>wave-particle duality</em>).

(iii) Werner Heisenberg — 1927 — formulated quantum mechanics in terms of matrices and proposed his famous <em>uncertainty principle</em>.

(v) Erwin Schrödinger — 1926 — applied wave mechanics to the electron in a hydrogen atom, showing that electrons exist in <em>orbitals </em>rather that orbits.

(iv) <em>Ernest Rutherford</em> — 1911 — proposed that atoms have most of their mass in a central nucleus (<em>nuclear atom</em>). Quantum mechanics had not yet been invented.

4 0
3 years ago
Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i
ankoles [38]
First, we need to get the value of Ka:

when Ka = Kw / Kb 

we have Kb = 1.8 x 10^-5 

and Kw = 3.99 x 10^-16  so, by substitution:

Ka = (3.99 x 10^-16) /  (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

                 NH4+  + H2O →NH3 +  H+
intial          0.013                       0         0 

change       -X                          +X      +X

Equ        (0.013-X)                      X         X

when Ka = [NH3][H+] / [NH4+] 

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X)  by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]
 
        = -㏒(5.35 x 10^-7)
        = 6.27
6 0
2 years ago
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