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3 years ago

Please hurry Participants in an informal discussion would most likely

Chemistry

2 answers:

Artist 52 [7]3 years ago

6 0

The answer is A informal is when you have a relaxed, friendly, or unofficial style, manner, or nature

Katyanochek1 [597]3 years ago

5 0

The answer is A. use casual, friendly conversation.

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If the temperature of a gas increased and the pressure remains the same what happens to the volume of gas?

podryga [215]

Answer:

the answer is B.

Explanation:

The pressure increases... plus i had this same question

pls give me brainliest

3 0

2 years ago

Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ

JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

$\frac{1}{2}$ Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is $\frac{1134.9}{4}$ = 283.725 kJ ⋅ mol − 1

4 0

3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

5 0

2 years ago

Consider the reaction given below.

Drupady [299]

Answer:

K = 0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.

Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

2ᵃ = 2 ⇒ a = 1

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K = 0.167 s⁻¹

6 0

3 years ago

Select the correct answer. What is the percent composition of silicon in silicon carbide (SiC)? A. 28% B. 50% C. 70% D. 142%

Nadya [2.5K]

Answer:

C. 70%

Explanation:

Atomic Mass of the silicon = 28 g.

Atomic mass of the Carbon = 12 g.

Total mass of the Silicon Carbide = 28 + 12

= 40 g.

Now, Using the formula.

% Composition = Mass of the silicon/Total mass of the compound × 100 %

= 28/40 × 100 %

= 70 %

Hence, % composition of the silicon in SiC is 70%

7 0

2 years ago