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3 years ago

Discuss whether two products are required for a chemical reaction. (NEED HELP ASAP)

Chemistry

1 answer:

Alex787 [66]3 years ago

6 0

Answer:

Yes.

Explanation:

"These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products."

There is a beginning product, and a reactant is needed in order for something to happen.

For example, according to Newton, something cannot happen until an exterior force comes and stops/pushes it.

Sorry if this is incorrect, I am just 4th grade :(

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A ball is thrown upward as shown. Gravity pulls (accelerates) the ball back down. Which of the following statements is true?

liraira [26]

Answer:

I think third one is the right answer

Explanation:

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3 years ago

Read 2 more answers

Be sure to answer all parts. Propane (C3H8) is a minor component of natural gas and is used in domestic cooking and heating. (a)

Leokris [45]

Answer:

For a: The balanced chemical equation is given below.

For b: The mass of carbon dioxide produced will be $1.07\times 10^3g$

Explanation:

For a:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor can be destroyed but it can only be transformed from one form to another form.

This law also states that the total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the given reaction, the balance chemical equation follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

All the substances are present in gaseous state.

For b:

By Stoichiometry of the reaction:

1 mole of propane gas produces 3 moles of carbon dioxide gas.

So, 8.11 moles of propane gas will produce = $\frac{3}{1}\times 8.11=24.33mol$ of carbon dioxide gas.

Now, calculating the mass of carbon dioxide using equation:

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

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 24.33 mol

Putting values in above equation, we get:

$24.33mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=1070.52g$

Hence, the amount of $CO_2$ produced in the given reaction and expressed in scientific notation is $1.07\times 10^3g$

3 0

4 years ago

55. Which Element has a larger radius- Rb or I?

Luden [163]

Rb has larger radius.

5 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

What mass (g) of potassium chloride is contained in 430.0 mL of a potassium chloride solution that has a chloride ion concentrat

fredd [130]

Answer:

6.178 g.

Explanation:

KCl is ionized in the solution as:

KCl → K⁺ + Cl⁻.

∴ [KCl] = [K⁺] = [Cl⁻] = 0.193 M.

Molarity is defined as the no. of moles of the solute per 1.0 L of the solution.

∵ M = [(mass/molar mass)KCl][(1000)/(V of the solution)].

∴ mass of KCl = (M)(molar mass of KCl)(V of the solution)/1000 = (0.193 M)(74.55 g/mol)(430.0 mL)/1000 = 6.178 g.

8 0

3 years ago