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hich of the following statements is not true for an exothermic reaction? Question 2 options: The products have a higher heat con

yKpoI14uk [10]

Answer:

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.

3 0

4 years ago

How does air flow when a low-pressure center occurs in the atmosphere?

Monica [59]

Answer:

A

Explanation:

It is so because the low air pressure create vacuum and the air from high pressure area move toward the low air pressure.

8 0

3 years ago

Love yall hope you doing good also anyone feel free to talk

statuscvo [17]

I’m doing good thanks

5 0

3 years ago

Read 2 more answers

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

What would happen to the gas pressure inside the cylinder if you do the following?

Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.

T₂ = 0.5 T₁. According to Gay-Lussac's law,

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}$

Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

$P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}$

P₂ in terms of the ideal gas equation is:

$P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}$

3 0

3 years ago

Consider this reaction 2Mg(s)+O2(g) ———> 2MgO(s) What volume (in milliners) of gas is required to react with 4.03 g Mg at STP

nydimaria [60]

The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml

calculation/

calculate the moles of Mg used

moles=mass/molar mass

moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles

by use of mole ratio of Mg:O2 from the equation which is 2:1

the moles 02=0.1679 x1/20.0829 moles

at STP 1 mole of a gas= 22.4 l

0.0895 moles=? L

by cross multiplication

=0.0895 moles x22.4 l/ 1 mole=1.8570 L

into Ml = 1.8570 x1000=1856 ml approximately to 1860

6 0

3 years ago