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Kisachek [45]
3 years ago
7

Given w = 0, an endothermic reaction has the following.

Chemistry
1 answer:
tamaranim1 [39]3 years ago
8 0

D) + ΔH and +ΔE

Given this is one of the answer choices

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Pls HELP!!!! Why does copper oxide appear on a penny faster when a penny is heated in a flame than when the penny is at room tem
balu736 [363]

The rate of reaction between oxygen and copper is higher at high temperature. The rate of reaction between oxygen and copper is lower at room temperature.

at high temperature copper penny will react with oxygen as shown below

2Cu(s)+O_{2}-->2CuO

The reaction does not occur easily at room temperature

We can say that the reaction between copper and oxygen is an endothermic reaction so the rate of reaction is high at high temperature.

The bond energy of oxygen molecules need energy to react with copper and form copper oxide.

5 0
3 years ago
Part 1. Determine the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm. Show your work.
zaharov [31]

If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.

An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given :

  • V = 2.4 L = 0.0024
  • P = 86126.25 Pa
  • T =  287 K
  • m = 0.622
  • R = 8.314

The ideal gas equation is given below.

n = PV/RT

n = 86126.25 x 0.0024 / 8.314 x 287

n = 0.622 / molar mass (n = Avogardos number)

Molar mass =  7.18 g

Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g

More about the ideal gas equation link is given below.

brainly.com/question/4147359

#SPJ1

4 0
2 years ago
It is difficult to measure the correct atomic radius. Why?
Paul [167]

Answer:

its kinda you just need to simplify

Explanation:

4 0
3 years ago
A first order reaction has rate constants of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0ºC and 20ºC, respectively. What is the value
Airida [17]

Answer:

D.  18,800 J/mol

Explanation:

We need to use the Arrhenius equation to solve for this problem:

k=Ae^{\frac{-E_a}{RT}, where k is the rate constant, A is the frequency factor, E_a is the activation energy, R is the gas constant, and T is the temperature in Kelvins.

We want to find the value of E_a, so let's plug some of the information we have into the equation. The gas constant we can use here is 8.31 J/mol-K.

At 0°C, which is 0 + 273 = 273 Kelvins, the rate constant k is 4.6*10^{-2}. So:

k=Ae^{\frac{-E_a}{RT}

4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}

At 20°C, which is 20 + 273 = 293 Kelvins, the rate constant k is 8.1*10^{-2}. So:

k=Ae^{\frac{-E_a}{RT}

8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}

We now have two equations and two variables to solve for. We just want to find Ea, so let's write the first equation for A in terms of Ea:

4.6*10^{-2}=Ae^{\frac{-E_a}{8.31*273}

A=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }

Plug this in for A in the second equation:

8.1*10^{-2}=Ae^{\frac{-E_a}{8.31*293}

8.1*10^{-2}=\frac{4.6*10^{-2}}{e^{\frac{-E_a}{8.31*273}} }e^{\frac{-E_a}{8.31*293}

After some troublesome manipulation, the answer should come down to be approximately:

Ea = 18,800 J/mol

The answer is thus D.

5 0
3 years ago
Which of the following always changes when a substance undergoes a chemical change?
Ronch [10]

Answer:

C

Explanation:

All chemical reactions will show a change in color.

8 0
3 years ago
Read 2 more answers
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