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3 years ago

What is the volume of 60 g of ether if the density of ether is 70 g/mL

Chemistry

1 answer:

NNADVOKAT [17]3 years ago

4 0

V = 60.0 g/ 0.70 g/mL = 85.7 mL Hope this helps! ;D

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olganol [36]

Answer:

Purple flowers

Explanation:

Usually the dominant allele is a capital letter.

From the question, purple flowers are dominant to white flowers while white are recessive.

PP = Purple

pp = White

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2 years ago

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An atom of the element iron has an atomic number of 26 and an atomic mass number of 56. if it is neutral, how many protons, neut

allsm [11]

<span> 26 protons, 30 neutrons, 26 electrons</span>

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Which statement is true for nuclear reactions but not for chemical reactions?

IrinaVladis [17]

Answer:maybe C

Explanation:

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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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3 years ago

One mole of copper equals about how many atoms?

Arlecino [84]

One mole of copper equals 6.02 × 10^23 atoms. The answer is letter C. This follows the Avogaro’s law wherein 1 mole of a substance is equal to 6.02 x 10^23 atoms, formula units or molecules. This is applicable to all substances.

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