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4 years ago

Epsom salt is mgso4·7 h2o. what does the chemical formula become when this hydrate is gently heated?

1 answer:

6 0

Epsom salt, magnesium sulphate, or MgSO4.7H2O comes in different hydrates.

The heptahydrate (.7H2O) comes in as Epsom salt.

It readily loses one water of crystallization, or by gentle heating, to become the hexahydrate, (.6H2O).

When heated to 150 ° C, it further lose water of crystallization to the monohydrate (.H2O)

Finally, when heated to 200 ° C, it becomes anhydrous, i.e. a powder that has no more water of crystallization.

(ref. Wiki)

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I need to know all the answers for this homework!!!!

maw [93]

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3 years ago

True or False: If you have 2 moles of water and 2 moles of carbon, you have the same number of molecules.

nikklg [1K]

True.
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3 years ago

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3 years ago

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A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 d

DENIUS [597]

Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

where $k_2$ = rate constant at temp $T_2$

$k_1$ = rate constant at temp $T_1$

$E_a$ = activation energy

R= gas constant

$T_1$ = temperature = $20^0C=(20+273)K=293K$

$T_2$ = temperature = $30^0C=(30+273)K=303K$

$ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]$

$ln \frac{k_{2}}{k_{1}}=1.15$

$\frac{k_{2}}{k_{1}}=3$

Thus rate increases 3 times on raising the temperature from 20degree to 30 degree

3 0

3 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

3 years ago