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Genrish500 [490]

3 years ago

5

What is stoichiometry used for

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