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2 years ago

How many atoms of oxygen are present in the reactants and products? 6CO2+6H2O-> C6H12O6+6O2

Chemistry

1 answer:

REY [17]2 years ago

6 0

6 Carbon atoms
18 Oxygen atoms
12 Hydrogen atoms

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Predict the product for this reaction.

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The answer is = LiCl ..............

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2 years ago

Which scientist was the first to use the telescope in astronomy? Copernicus Newton Kepler Galileo

coldgirl [10]

The scientist that was the first to use the telescope in astronomy was Newton

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2 years ago

What adaptations allow a camel and a cactus to survive in warm environments?

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2 years ago

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Cl2 and N2 react according to the following equation 3Cl2(g) + N2(g) rightarrow 2NCl3(g) If 4 L of a stoichiometric mixture of c

melamori03 [73]

Answer:

Volume of NCl3 is 3L

Explanation:

Avogadro states: All gases at the same volume under temperature and pressure constant have the same number of moles.

The chemical equation is:

3Cl2(g) + N2(g) → 2NCl3(g)

Where 3 moles of chlorine reacts with 1 mole of nitrogen to produce 2 moles of NCl3.

But using Avogadros law we can say:

3L of chlorine and 1L of nitrogen produce 2L of Nitrogen trichloride.

3L of chlorine and 1L of nitrogen: 4L (The stoichiometric mixture)

That means, volume of NCl3 produced is 3L

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2 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

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2 years ago