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1 day ago

Calculate the mass of hydrazine (N2H4) that contains a billion (1.000 x 10^9) hydrogen atoms.

2 answers:

8 0

To find the mass of hydrazine containing 1 billion hydrogen atoms, we state that , thus the mass is $1.328\times 10^{-14}g$

Explanation:

The molar mass for hydrazine is 32 g/mol

. Given that 1 mole of hydrazine consists of 2 moles of nitrogen atoms and 4 moles of hydrogen atoms, if we need the number of hydrogen atoms = $1.000\times 10^9$

Based on the mole concept, 1 mole of a substance contains Avogadro's number of molecules.

Thus, 1 mole of hydrazine will contain a certain number of hydrogen atoms. Hence, the mass for 1 billion hydrogen atoms in hydrazine equals 32 grams.

Consequently, for 1 billion hydrogen atoms, the mass of hydrazine equates to $\frac{32g}{2.4088\times 10^{24}}\times 1.000\times 10^9=1.328\times 10^{-14}g$

lorasvet [2.6K]1 day ago

3 0

The mass of hydrazine required to contain 1 billion hydrogen atoms is 1.3285 × 10⁻¹⁴ g

Explanation:

Given that 32 g of hydrazine (equivalent to 1 mole) contains 4 × 6.022 × 10²³ hydrogen atoms, we can set up the equation: X g of hydrazine holds 1 × 10⁹ hydrogen atoms. Thus, X = (32 × 1 × 10⁹) / (4 × 6.022 × 10²³) leads us to conclude that X = 1.3285 × 10⁻¹⁴ g of hydrazine.

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In a chemical reaction, 300 grams of reactant A are combined with 100 grams of reactant B. Both A and

KiRa [2857]

I expect the product's mass to be 400 grams. This belief stems from the law of conservation of mass, which states that mass can neither be created nor destroyed. In a sealed system, the mass of the reactants equals the mass of the products. Therefore, since the total mass of the reactants is 400 grams, the resulting mass of the products must also be 400 grams

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8 0

7 days ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

lions [2782]

Answer: The molecular formula for the specified organic compound is $C_{18}H_{20}O_2$

Explanation:

The combustion reaction of a hydrocarbon comprising carbon, hydrogen, and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' denote the subscripts of Carbon, hydrogen, and oxygen respectively.

The information provided includes:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

From our knowledge:

Molar mass of carbon dioxide is 44 g/mol

Molar mass of water is 18 g/mol

For determining the amount of carbon:

In carbon dioxide weighing 44 g, 12 g of carbon is found.

Hence, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ grams of carbon will be found.

For finding the mass of hydrogen:

In water weighing 18 g, 2 g of hydrogen can be found.

Thus, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ grams of hydrogen will be present.

Mass of oxygen in the compound is given by (13.42) - (10.80 + 1.00) = 1.62 g

To derive the empirical formula, the following steps must be followed:

Step 1: Convert the indicated masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the ratio of moles of the respective elements.

To find the mole ratio, each mole value is divided by the smallest amount of moles calculated, which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3:Using the mole ratios as subscripts.

The ratio of C: H: O = 9: 10: 1

Therefore, the empirical formula for the mentioned compound is $C_9H_{10}O$

To ascertain the molecular formula, it is necessary to find the valency, which is multiplied by each element to derive the molecular formula.

The equation to determine the valency is:

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

Given the data:

Molecular formula mass = 268.34 g/mol

Empirical formula mass = 134 g/mol

Substituting the values into the aforementioned equation yields:

$n=\frac{268.34g/mol}{134g/mol}=2$

By multiplying this valency with each element's subscripts from the empirical formula, the results are:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Consequently, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

9 days ago

How many grams of AlF3 are in 2.64 moles of AlF3?

Anarel [2728]

To determine the mass of AlF3 in 2.64 moles of AlF3, we use the formula: mass = moles x molar mass, which results in 221.76 grams of AlF3.

3 0

22 days ago

A certain chemical reaction releases 362.kj of heat energy per mole of reactant consumed. Suppose some moles of the reactant are

lions [2782]

Answer:

The temperature increase of the calorimeter, which is missing in the problem, is necessary for the calculation.

Explanation:

Since the temperature rise (X) is unspecified, we'll express the calculation in terms of X, and demonstrate with an example value.

1) Calorimeter details:

Temperature increase: X °C

Heat capacity ratio: 4.87 J / 5.5 °C (given)

Energy absorbed by calorimeter at X °C rise:

(4.87 J / 5.5 °C) × X

2) Reaction data:

Heat released: 362 kJ per mole of reactant

Number of moles consumed: n

Total energy from reaction:

362 kJ/mol × 1000 J/kJ × n = 362,000 n J

3) Using energy conservation, assuming no heat loss to surroundings, the energy from the reaction equals the energy absorbed by the calorimeter:

362,000 n = (4.87 J / 5.5 °C) × X

Solving for n gives:

n = [(4.87 / 5.5) × X] / 362,000

n = 0.000002446 × X

This means for each degree Celsius rise in calorimeter temperature, 0.000002446 moles of reactant were consumed.

Example:

If the calorimeter temperature increases by 100 °C, then:

n = 0.000002446 × 100 = 0.0002446 mol

6 0

1 month ago

What is the volume in ml of a 1.11 carat diamond, given the density of diamond is 3.51 g/ml?

KiRa [2857]

The answer is that the diamond's volume measures 0.063 ml.
With a density of d(diamond) = 3.51 g/ml, and a mass of m(diamond) = 1.11 carat, with 1 carat being equivalent to 0.2 grams, we convert m(diamond) to grams: m(diamond) = 1.11 carat·0.2 gram/carat, which gives m(diamond) = 0.222 g.
To find the volume: V(diamond) = m(diamond) ÷ d(diamond), which results in V(diamond) = 0.222 g ÷ 3.51 g/ml = 0.063 ml.

3 0

24 days ago