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Genrish500 [490]
3 years ago
5

What is stoichiometry used for

Chemistry
2 answers:
Julli [10]3 years ago
4 0

Answer:

Explanation:

they measures these quantitative relationships. But not dating relationships, but the amount of products and reactants that are produced or needed in a given reaction.

Dennis_Churaev [7]3 years ago
3 0

Answer:

Stoichiometry measures these quantitative relationships, and is used to determine the amount of products and reactants that are produced or needed in a given reaction.

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What form of energy causes an ice cube to melt
zalisa [80]

Answer:

Thermal

Explanation:

When something heats up its atoms become further apart making it melt when something heats up or gets heat or loses heat that's thermal energy

6 0
3 years ago
Which may be true for a substance that absorbs energy?
irinina [24]
It has different moleculs sorry I am guessing
8 0
2 years ago
Given the equation I = Q/t, solve for t.
larisa86 [58]

Answer: t=\frac{Q}{I}

Explanation:

I=\frac{Q}{t}

Multiply by t on both sides.

t*I=\frac{Q}{t}*t

tI=Q

Now divide by I to isolate t.

\frac{tI}{I}=\frac{Q}{I}

t=\frac{Q}{I}

5 0
3 years ago
Is copper ii sulphate (solution) blue or green?
slavikrds [6]
Copper II sulfate solution is blue.
8 0
3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

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

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
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