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2 years ago

Calculate the density of an object that has a mass of 43 g and a volume of 56.0 ml.

Chemistry

1 answer:

Yanka [14]2 years ago

5 0

I will try to do it but not more than 70.4

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What type of solution is formed when steam from boiling water evaporates into the air

Hoochie [10]

That specific kind is called gas.

When you put water in a freezing temprature it becomes solid.
When you boil water it becomes gas

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3 years ago

1. Intermolecular forces do not have any effect on determining the physical properties of substances such as surface tension and

Scilla [17]

Answer:

1)false

2) false

3)True

4)True

5) false

6)false

7) true

8) d

9)false

10) b

5 0

2 years ago

The bonds of the product store 27 KJ more energy than the bonds of the reactants how is energy conserved during this reaction

Blababa [14]

If the bonds of the product store 27 KJ more energy than the bonds of the reactants, It means the surroundings absorb 27 kj of energy from the reaction system Hence, Option (D) is the correct answer

<h3>What is the Exothermic reaction ?</h3>

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise.

The bonds of the product store 27 KJ more energy than the bonds of the reactants, It means that energy has been absorbed by the surrounding as the product formed is more stable due to more stronger bond

This can be inferred from more stored energy with in the bonds and Thus, It is a exothermic reaction.Hence, Option (D) is the correct answer

Learn more about exothermic reaction here ;

brainly.com/question/9799465

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6 0

1 year ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}