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3 years ago

Which element (A, B, or C) is most likely a metalloid

Chemistry

2 answers:

Elza [17]3 years ago

5 0

Answer:

its C

Explanation:

BigorU [14]3 years ago

3 0

Answer: B

Explanation:

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1. Define the unit 'mole'

serious [3.7K]

Answer:

Below

Explanation:

The mole, symbol mol, is the SI base unit of amount of substance. The quantity amount of substance is a measure of how many elementary entities of a given substance are in an object or sample

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2 years ago

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The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.751 g sample of e

fredd [130]

Answer: The empirical formula for the given compound is $C_{4}H_{10}O$

Explanation:

The chemical equation for the combustion of ether follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=4.159g$

Mass of $H_2O=2.128g$

Mass of sample = 1.751 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 4.159 g of carbon dioxide, $\frac{12}{44}\times 4.159=1.134g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.128 g of water, $\frac{2}{18}\times 2.128=0.236g$ of hydrogen will be contained.

Mass of oxygen in the compound = (1.751) - (1.134 + 0.236) = 0.381 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.134g}{12g/mole}=0.0945moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.236g}{1g/mole}=0.236moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.381g}{16g/mole}=0.0238moles$

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 0.0238 moles.

For Carbon = $\frac{0.0945}{0.0238}=3.97\approx 4$

For Hydrogen = $\frac{0.236}{0.0238}=9.91\approx 10$

For Oxygen = $\frac{0.0238}{0.0238}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is $C_{4}H_{10}O$

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3 years ago

Calculate the molarity of 32.1 g of MgS in 841 mL of solution.

Marianna [84]

Molarity = mol / liter = 0.569 / 0.841 = 0.677

You convert 841 mL of solution to 0.841 L
You convert 32.1g of MgS to mol by dividing by molar mass of MgS (which is 56.4g)

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