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ira [324]
3 years ago
14

The picture is a waning gibbous. What phase will the moon be two weeks later?

Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

d

Explanation:

Oksana_A [137]3 years ago
4 0

Answer:

D

Explanation:

Because it is l o l

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A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
3 years ago
If have a volume of 18 L of a gas at a temperature of 272 K and a pressure of 90 atm, what will be the pressure of the gas if ra
Solnce55 [7]

Answer:

P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)

Explanation:

Given:

P₁ = 90 atm                    P₂ = ?

V₁ = 18 Liters(L)              L₂ = 12 Liters(L)      

=> decrease volume => increase pressure

=> volume ratio that will increase 90 atm is (18L/12L)                                                                  

T₁ = 272 Kelvin(K)          T₂ = 274 Kelvin(K)

=>  increase temperature => increase pressure

=> temperature ratio that will increase 90 atm is (274K/272K)

n₁ = moles = constant    n₂ = n₁ = constant

P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)

By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.

3 0
2 years ago
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e
TEA [102]

<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

<u>Explanation:</u>

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = -PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


6 0
3 years ago
What is true of a substance with a lot of mass?
fomenos

Answer:

A. it contains a lot of matter

5 0
2 years ago
Read 2 more answers
Why can't the reaction, ZnCl2 + H2 → Zn + 2HCI, occur naturally?
Llana [10]

Answer:

It is fairly obvious that zinc metal reacts with aqueous hydrochloric acid! The bubbles are hydrogen gas. ... In fact, electrons are being transferred from the zinc atoms to the hydrogen atoms (which ultimately make a molecule of diatomic hydrogen), changing the charges on both elements.

Explanation:

7 0
3 years ago
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