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3 years ago

A compound is found to contain 6.1% hydrogen and 93.9% oxygen. Find it’s empirical formula.

Chemistry

1 answer:

Kay [80]3 years ago

3 0

Answer:

CH.

Explanation:

If the mass of the sample is 1.0 g, this means that it contains 6.1 g hydrogen and 93.9 g oxygen.

We can obtain the no. of moles by dividing by its atomic masses.

The no. of moles of hydrogen = mass / atomic mass = (6.10 g)/(1.00 g/mol) = 6.10 mol.

The no. of moles of oxygen = mass / atomic mass = (93.90 g)/(16.00 g/mol) = 5.86 mol.

We can obtain the empirical formula of the compound by dividing both no. of moles by the lowest no. of moles (5.86)

∴ The ratio of H:O is (1.04: 1.00) which is mainly 1.0: 1.0.

<em>So, the empirical formula of the compound is CH.</em>

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Is milk a solution, colloid, suspension, or a compound?

allsm [11]

Milk is a colloid system

3 0

3 years ago

A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3

Sati [7]

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C

(58 g . 0.100) / 0.580 = 10 g of H

48 g of C / 12 g/mol = 4 mol

10 g of H / 1g/mol = 10 moles

7 0

3 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

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

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

4 0

3 years ago

Determine the mass (in grams) of 8.02 x 1024 atoms of mercury (Hg).

Virty [35]

<h3>Answer:</h3>

2670 g Hg

<h3>General Formulas and Concepts: </h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation: </h3>

<u>Step 1: Define</u>

8.02 × 10²⁴ atoms Hg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 8.02 \cdot 10^{24} \ atoms \ Hg(\frac{1 \ mol \ Hg}{6.022 \cdot 10^{23} \ atoms \ Hg})(\frac{200.59 \ g \ Hg}{1 \ mol \ Hg})$
Divide/Multiply: $\displaystyle 2671.42 \ g \ Hg$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2671.42 g Hg ≈ 2670 g Hg

3 0

3 years ago

The decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C is independent of the concentration of amm

finlep [7]

Answer:

Order zero

Explanation:

Let's consider the decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C.

2 NH₃(g) → N₂(g) + 3 H₂(g)

The generic rate law is:

rate = k × [NH₃]ⁿ

where,

rate: reaction rate

k: rate constant

n: reaction order

When n = 0, we get:

rate = k × [NH₃]⁰ = k

As we can see, when the reaction order with respect to ammonia is zero, the reaction rate is independent of the concentration of ammonia.

7 0

3 years ago