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3 years ago

A 120.5 g sample of water loses -7525 J of energy. What was the

Chemistry

1 answer:

dusya [7]3 years ago

5 0

Answer:14.93

Explanation:heat gained=heat lost

120.5 ×4.184×change=7525

7525÷504.172

Ans14.925

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For each store,what is the ratio of the number of cans to the price:24 packs (2 for 9);12 packs (4 for 10);12 packs (3 for 9)

adelina 88 [10]

$4.50 for the 24 pack, $2.50 for the 4 for 10, 12 pack and for the last one it is $3.00

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3 years ago

How many moles of hydrogen are needed to produce 8 moles of methane CH4? C + H2-> CH4

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Answer:

20 mol H₂

General Formulas and Concepts:

Chem

Stoichiometry

Explanation:

Step 1: Define

RxN: C + H₂ → CH₄

8 moles CH₄

Step 2: Balance RxN

C + 2H₂ → CH₄

Step 3: Stoichiometry

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Step 4: Check

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2 years ago

Determine the pH of the following base solutions. (Assume that all solutions are at 25°C and the ion-product constant of water,

alexira [117]

Answer:

a) The pH of the solution is 12.13.

b) The pH of the solution is 12.17.

Explanation:

Ionic product of water = $K_w=1.01\times 10^-{14}$

$K_w=[H^+][OH^-]$

$1.01\times 10^-{14}=[H^+][OH^-]$

Taking negative logarithm on both sides:

$-\log[1.01\times 10^-{14}]=(-\log [H^+])+(-\log [OH^-])$

The pH is the negative logarithm of hydrogen ion concentration in solution.

The pOH is the negative logarithm of hydroxide ion concentration in solution.

$13.99=pH+pOH$

a) $1.39\times 10^{-2} M$ of NaOH.

Concentration of hydroxide ions:

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

So, $[OH^-]=1\times [NaOH]=1\times 1.39\times 10^{-2} M=1.39\times 10^{-2} M$

$pOH=-\log[1.39\times 10^{-2} M]=1.86$

$13.99=pH+pOH$

$13.99=pH+1.86$

pH=13.99-1.86=12.13

b) $0.0051 M$ of NaOH.

Concentration of hydroxide ions:

$Al(OH)_3(aq)\rightarrow Al^{3+}(aq)+3OH^-(aq)$

So, $[OH^-]=3\times [Al(OH)_3]=3\times 0.0051 M=0.0153 M$

$pOH=-\log[0.0153 M]=1.82$

$13.99=pH+pOH$

$13.99=pH+1.82$

pH=13.99-1.82=12.17

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