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3 years ago

40 g of calcium reacts with 71 g of chlorine to produce _____ g of calcium chloride

Chemistry

2 answers:

grin007 [14]3 years ago

8 0

The answer is 111g of calcium chloride

Andreas93 [3]3 years ago

5 0

Answer: 111 g

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles of calcium}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{40g}{40g/mol}=1mole$

$\text{Number of moles of chlorine}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{71g}{71g/mol}=1mole$

$Ca+Cl_2\rightarrow CaCl_2$

1 moles of calcium react with 1 mole of chlorine to give 1 mole of calcium chloride.

Mass of calcium chloride = $moles\times {\text {molar mass}}=1mol\times 111g/mol=111g$

Thus 111 g of calcium chloride is produced.

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For the skeletal chemical equation BF3(g) + NaH(s) → B2H6(g) + NaF(s) what is the coefficient of B2H6 in the balanced equation?

KonstantinChe [14]

Answer:

Coefficient in front of the $B_2H_6$ in the balanced equation - 1

Explanation:

The unbalanced Chemical equation is shown below as:-

$BF_3+NaH\rightarrow B_2H_6+NaF$

On the left hand side,

There are 1 boron atom and 3 fluorine atoms and 1 sodium and hydrogen atoms.

On the right hand side,

There are 2 boron atoms and 6 hydrogen atoms and 1 sodium and fluorine atoms.

Thus,

leftside, $BF_3$ must be multiplied by 2 to balance boron and right side, $NaF$ must be multiplied by 6 to balance fluorine. Left side, $NaH$ must be multiplied by 6 to balance sodium and hydrogen atoms.

Thus, the balanced reaction is:-

$2BF_3+6NaH\rightarrow B_2H_6+6NaF$

Coefficient in front of the $B_2H_6$ in the balanced equation - 1

6 0

3 years ago

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Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation

nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ$ ..[2]

$H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

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2 years ago

What happens when metamorphic rock becomes igneous rock?

Nookie1986 [14]

D. metamorphic rock melts and crystallizes to become igneous rock

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3 years ago

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Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this

maw [93]

Answer: d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

$2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H=+1411.1kJ$

Reversing the reaction, changes the sign of $\Delta H$

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$

$\Delta H=-1411.1kJ$

On multiplying the reaction by $\frac{1}{2}$ , enthalpy gets half:

$0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)$ $\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol$

Thus the enthalpy change for the given reaction is -705.55kJ

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3 years ago

Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and

tatyana61 [14]

Answer: The equilibrium concentration of $PCl_5$ is 1.285 M.

Explanation:

The chemical equation for the decomposition of phosphorus pentachloride follows:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression for equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We are given:

$K_c=0.042$

$[PCl_3]=0.18M$

$[Cl_2]=0.30M$

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

$0.042=\frac{0.18\times 0.30}{[PCl_5]}$

$[PCl_5]=1.285$

Hence, the equilibrium concentration of $PCl_5$ is 1.285 M.

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3 years ago