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strojnjashka [21]
4 years ago
7

Bohr model for silicon

Chemistry
1 answer:
Lunna [17]4 years ago
5 0
Hope this answer helps you

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2C2H2(g) + 502(g) → 4C02(g) + 2H20(g) reaction type?
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The answer is combustion.

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2 years ago
The following reaction is exothermic. C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g)C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g) Predict the effect (shi
elena55 [62]

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According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.

Explanation:

C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)

We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.

According to Le Chatelier's principle,

1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.

2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.

3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.

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3 years ago
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Will mark the brainiest later for correct answers! Please show work.
Elodia [21]

Answer:

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

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}

a. moles in 14.08 g of C_{12}H_{22}O_{11} = \frac{14.08g}{342.3g/mol}=0.04113moles

molecules in 14.08 g of C_{12}H_{22}O_{11} = 0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}

b. moles in 17.75 g of NaCl = \frac{17.75g}{58.5g/mol}=0.3034moles

molecules in 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

formula units 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

c. moles in 20.06 g of  CuSO_4.5H_2O= \frac{20.06g}{249.68g/mol}=0.08034moles

formula units in 20.06 g of  CuSO_4.5H_2O= 0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}

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