All categories

3 years ago

Which of the following statements is true?

Chemistry

2 answers:

zalisa [80]3 years ago

6 0

It's A, tThe figure is a molecule and an element.

zhuklara [117]3 years ago

4 0

The answer is A. The figure is a molecule and an element. You're welcome

You might be interested in

Some molecules in air contain . Explain how atoms from the air can become part of human cells

Tcecarenko [31]

The atoms in air are combined with oxygen present in the air and when we inhale the air, oxygen is moved into the body then atoms in oxygen are carried by red blood cells the blood is pumped to the lungs and when oxygen is transported to the body organ, the atoms in lungs become the part of human cells.

5 0

2 years ago

Write the balanced half equation of iron 2 and permanganate in a solution of acid. Show all of your work.

Aleks [24]

Answer:

5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

Explanation:

Fe⁺² + MnO₄⁻ + H⁺ => Mn⁺² + Fe⁺³ + H₂O

5(Fe⁺² => Fe⁺³ + 1e⁻) => 5Fe⁺² => 5Fe⁺³ + 5e⁻

MnO₄⁻ + 5e⁻ => Mn⁺² => MnO₄⁻ + 8H⁺ + 5e⁻ => Mn⁺² + 4H₂O

=> 5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O

6 0

3 years ago

Adding energy to a solid bar of gold may result in which of the following outcomes

____ [38]

An increase in motion and less attraction between particles

4 0

2 years ago

Oksi-84 [34.3K]

Answer:

900 mg

Explanation:

4 0

3 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

3 years ago