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3 years ago

Which compound is an electrolyte?

2 answers:

6 0

Your answer is (4) methanoic acid, more commonly known as the formic acid. It is a weak electrolyte that dissolves into H^+ and COOH^-

Masteriza [31]3 years ago

4 0

The compound that is an electrolyte is Methanoic acid. Because it ionizes in water, which makes it an electrolyte.

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An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is

aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x ( $\frac{1 L}{1000 mL}$ ) = .236 L

Pressure= 740 mm Hg x ( $\frac{1 atm}{760 mm Hg}$ )= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= $\frac{mass}{moles}$ or MM= $\frac{m}{n}$ . They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n= $\frac{PV}{RT}$ , where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n= $\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}$

n= 0.009 mol

Let's look back at MM= $\frac{m}{n}$ and plug in what we know.

MM= $\frac{0.443 g}{0.009 mol}$

MM= 49.2 g/mol

3 0

3 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

3 0

2 years ago

Calculate the molecular masses of C4 H10 atoms molecules or ionic formulae

Basile [38]

C4H10 is an alkane (propane to be precise)

The relative formula mass of a molecule of propane is

(12x4) + (1x10) = 58

because a carbon atom has an atomic weight of 12 and a hydrogen atom has an atomic weight of 1.

8 0

2 years ago

Use the following equation to answer question 1-4. Make sure you balance first.

worty [1.4K]

<h3>Answer:</h3>

5.2 mol H₂O

<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>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

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

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})$
Multiply/Divide: $\displaystyle 5.2 \ mol \ H_2O$

4 0

3 years ago

Telephone signals are often transmitted over long distances by microwaves. what is the frequency of microwave iradiation with a

Alborosie

<span>c = speed of light = 3.00 x 10^5 km/s = 3.00 x 10^8 m/s
Î» = wavelength of the microwave radiation = 3.50 cm = 0.035 m
f = frequency (in Hertz) = to be determined
f = c/Î» = 3.00 x 10^8 m/s / 0.035 m
f = 8.57 x 10^9 Hz Frequency</span>

6 0

4 years ago

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