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2 years ago

Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).

Chemistry

1 answer:

vova2212 [387]2 years ago

3 0

Energy absorbed by Iron block E (iron) = 460.5 J

Energy absorbed by Copper block E (Copper) = 376.8 J

<u>Explanation:</u>

To find the heat absorbed, we can use the formula as,

q = m c ΔT

Here, Mass = m = 10 g = 0.01 kg

ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C

c = specific heat capacity

c for iron = 460.5 J/kg K

c for copper = 376.8 J/kg K

Plugin the values in the above equation, we will get,

q (iron) = 0.01 kg × 460.5 J/kg K × 100 K

= 460.5 J

q (copper) = 0.01 kg × 376.8 J/kg K × 100 K

= 376.8 J

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Gas y effuses half as fast as o2. what is the molar mass of gas y?

devlian [24]

To solve this question, we will use Graham's law which states that:
(R1 / R2) ^ 2 = M2 / M1 where
R1 and R2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.
From the periodic table, we can calculate the molar mass of O2 as follows:
molar mass of O2 = 2*16 = 32 grams

Therefore we have:
R1 / R2 = Ry / RO2 = 1/2
M1 is My we want to get
M2 is molar mass of O2 = 32 grams
Substitute in the above equation to get the molar mass of y as follows:
(1/2) ^2 = (32/My)
1/4 = 32/My
My = 32*4 = 128

Therefore, molar mass of gas y = 128 grams

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3 years ago

The shininess of an element is also referred to as "metallic luster." which of the following elements is most likely to have a m

lisov135 [29]

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2 years ago

Using microorganisms to break down pollutants into nontoxic organic matter is called? a. bioremediation b. phytoremediation c. b

Brilliant_brown [7]

Answer:

bioremediation

Explanation:

3 0

2 years ago

3. A reaction produces 14.2 grams of a product. The theoretical yield of that product

nlexa [21]

A reaction that produces 14.2 grams of a product and the theoretical yield of that product is 17.1 grams is true for the following statements :

The percent yield of the product is 83.0%

The actual yield of the product is 14.2 grams.

<h3>Percentage Yield:</h3>

Percent yield is the percent ratio of actual yield to the theoretical yield.

Mathematically,

percent yield = actual yield / theoretical yield x 100%

actual yield = 14.2 grams

theoretical yield = 17.1 grams

percentage yield = 14.2 / 17.1 × 100%

percentage yield = 83.0409356725 %

percentage yield = 83.0 %

Therefore,

The percent yield of the product is 83.0%

The actual yield of the product is 14.2 grams.

learn more on percentage yield here; brainly.com/question/4180677

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2 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

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2 years ago