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yawa3891 [41]
3 years ago
7

How much heat will be released when 10.0 g of hydrogen peroxide decomposes according to the following reaction: 2H2O2(l) 2H2O(l)

+ O2(g) ΔH = -190 kJ
A. 26 kJ
B. 22 kJ
C. 28 kJ
D. 24 kJ
Chemistry
1 answer:
kramer3 years ago
5 0
C. 28 KJ

AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2

0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
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fluoride ion with a charge of -1

Explanation:

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The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is 1.1\times 10^6 M.

HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)

This value indicates that

A. CN^- is a stronger base than ONO^-

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is ONO^-

D. The conjugate acid of CN- is HCN

Answer: A. CN^- is a stronger base than ONO^-

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When K_{p}>1; the reaction is product favoured.

When K_{p}; ; the reaction is reactant favored.

When K_{p}=1; the reaction is in equilibrium.

As, K_p>>1, the reaction will be product favoured and as it is a acid base reaction where HONO acts as acid by donating H^+ ions and CN^- acts as base by accepting H^+

Thus HONO is a strong acid thus ONO^- will be a weak conjugate base and CN^- is a strong base which has weak HCN conjugate acid.

Thus the high value of K indicates that CN^- is a stronger base than ONO^-

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What are some sources of non-point source water pollution?
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Excess fertilizers, herbicides and insecticides from agricultural lands and residential areas.

Oil, grease and toxic chemicals from urban runoff and energy production.

Sediment from improperly managed construction sites, crop and forest lands, and eroding streambanks.

3 0
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Answer:

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Step-by-step explanation:

Before the ball is dropped, it has a <em>potential energy </em>

PE = mgh

PE = 0.2 × 10 × 9.8

PE = 19.6 J

Just before the ball hits the ground, the potential energy has been converted into kinetic (<em>mechanical</em>) energy.

KE = 19.6 J

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3 years ago
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