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3 years ago

Calcium chloride contains only calcium and chloride. What is the formula for this compound?

Chemistry

2 answers:

nataly862011 [7]3 years ago

4 0

Calcium chloride contains only one calcium and 2 chloride atom thus it's ionic formula would be CaCl2.

inna [77]3 years ago

4 0

Answer:

CaCI2

Explanation:

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galina1969 [7]

Answer:

Explanation:

C. energy being transferred from wind

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3 years ago

Please help Asap………………

vladimir2022 [97]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

M. element<br>cenabrecht<br>0.5 mele of an

drek231 [11]

Answer:

Number of moles (n)=

molecular weight

weight

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3 years ago

What do you put in the calculator to get the molar mass of Mg3P2

olga55 [171]

Input the atomic masses of Mg and P to give 134.84g/mol

Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

P = 30.97g/mole

Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

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Molar mass brainly.com/question/2861244

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6 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago