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3 years ago

15

Equal masses of two substances absorb the same amount of heat. The temperature of a substance A increases twice as much as subst

ance B. Which substance has a higher specific heat capacity?
A. Substance A
B. Substance B

1 answer:

katrin [286]3 years ago

7 0

Letter A because the temperature of a substance A increases "Twice as much".

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

<u>Initial:</u> 0.275

<u>At eqllm:</u> 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

What us the % composition of oxygen (O) in iron (III) hydroxide, Fe(OH)<br> 3?

S_A_V [24]

Answer:

44.91% of Oxygen in Iron (III) hydroxide

Explanation:

To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:

<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>

<em />

<em>Molar mass Fe(OH)3 and oxygen:</em>

1Fe = 55.845g/mol*1 = 55.845

3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen

3H = 1.008g/mol*3 = 3.024

55.845 + 48.00 + 3.024 =

106.869g/mol is molar mass of Fe(OH)3

% Composition of oxygen is:

48.00g/mol / 106.869g/mol * 100 =

<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>

7 0

2 years ago

You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2

shusha [124]

Ho123 right I’m sorry of I’m wrong

7 0

2 years ago

2. Given the unbalanced equation:

zheka24 [161]

2 Na + 2 H2O → 2 NaOH + H2 (balanced equation)
The answer would be 2, since 2 in the coefficient of both Na and NaOH

7 0

2 years ago

How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)

7nadin3 [17]

Answer:

The answer to your question is V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

CaH₂ = 40 + 2 = 42 g

42 g ------------------ 1 mol

2.65 g -------------- x

x = (2.65 x 1)/42

x = 0.063 moles

2.- Calculate the moles of NH₃

2 moles of NH₃ --------------- 3 moles of CaH₂

x --------------- 0.063 moles

x = (0.063 x 2) / 3

x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

= 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

PV = nRT

-Solve for V

V = nRT / P

-Substitution

V = (0.042)(0.082)(296) / 3.2

-Simplification

V = 1.019 / 3.2

-Result

V = 0.32 L

7 0

2 years ago