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alexandr402 [8]
3 years ago
13

Calculate the heat (in J) required to raise the temperature of 3.02 g of solid silver from 22.3 degrees Celsius so solid silver

at 47.0 degrees Celsius?
Chemistry
1 answer:
Furkat [3]3 years ago
7 0

Answer: 17.9 joules

Explanation:

The quantity of Heat Energy (Q) required to heat solid silver depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = ?

Mass= 3.02g

C = 0.24J/g°C

Φ = (Final temperature - Initial temperature)

= 47°C - 22.3°C = 24.7°C

Then, Q = MCΦ

Q = 3.02g x 0.240J/g°C x 24.7°C

Q = 17.9 Joules

Thus, 17.9 joules of heat isrequired.

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Blababa [14]

Answer:

A. 4.5 mol Mg(OH)₂

B. 6 mol NaOH

Explanation:

Let's consider the following balanced equation.

Mg(NO₃)₂ + 2 NaOH ⇒ Mg(OH)₂ + 2 NaNO₃

PART A

The molar ratio of NaOH to Mg(OH)₂ is 2:1. The moles of Mg(OH)₂ produced from 9 moles of NaOH are:

9 mol NaOH × 1 mol Mg(OH)₂/2 mol NaOH = 4.5 mol Mg(OH)₂

PART B

The molar ratio of NaOH to NaNO₃ is 2:2. The moles of NaOH needed to produce 6 moles of NaNO₃ are:

6 mol NaNO₃ × 2 mol NaOH/2 mol NaNO₃ = 6 mol NaOH

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3 years ago
To determine the concentration of 20.00 mL of an unknown solution of a monoprotic acid, it is titrated with 0.1093 M sodium hydr
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Answer:

d) 0.1202 M

Explanation:

Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M

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A balloon contains 7. 2 L of helium. The pressure is reduced to 2. 00 atm and the balloon expands to occupy a volume of 25. 1 L.
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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
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Answer:

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A + B + C → D + E

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aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

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Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

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Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

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a = 2

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R=k[A]^2[B]^0[C]^1

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k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

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