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2 years ago

Which formula represents the compound commonly known as phosphine with iupac name phosphorus trihydride? ph ph2 ph3 ph4

Chemistry

1 answer:

zubka84 [21]2 years ago

6 0

C. PH3 represents a compound commonly known as phosphine, whose IUPAC name is phosphorus trihydride.

The electronegativity of PH3 found in the Periodic Table of the Period attracts covalent electron pairs and creates covalent bonds. However, because the electrons are not bound, asymmetrical rate distribution occurs. Therefore, PH3 is a polar molecule with a non-polar covalent bond and currently has no polar bond.

<h3 /><h3>What defines a covalent bond?</h3>

A covalent bond consists of sharing one or more electron pairs between two atoms. These electrons are attracted to two nuclei at the same time. Covalent bonds are formed when the difference in electronegativity between two atoms is too small for electron transfer to form ions.

Click here for more information on covalent bonds brainly.com/question/12732708

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Getting empirical formula, but they also gave molar mass too.

-Dominant- [34]

Answer:

Empirical formula is PNCl₂

Explanation:

Percent composition means that 100 g of compound has x g of each element.

In 100 g of compound x, we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl.

So, let's make some rules of three:

In 100 g of compound we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl

In 579.43 g of compound we have:

(579.43 . 26.73) / 100 = 155 g of P

(579.43 . 12.09) / 100 = 70 g of N

(579.43 . 61.18) / 100 = 354 g of Cl

Let's convert the mass of the elements in moles.

155 g of P / 30.97 g/mol = 5 P

70 g / 14 g/mol = 5 N

354 g / 35.45 g/mol = 10 Cl

7 0

4 years ago

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Meghan (CH4) burns in oxygen to produce carbon dioxide and water.What reaction is occurring here?

Serga [27]

Combustion. You said CH4 'burns' in the question which is another word for combustion.

6 0

4 years ago

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What is the ph of a solution of 0.50 m acetic acid?

frosja888 [35]

You need to use the Ka for the acetic acid and the equilibrium equation.

Ka = 1.85 * 10^ -5

Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)

Ka = [CH3COO-][H+] / [CH3COOH]

Molar concentrations at equilibrium

CH3COOH CH3COO- H+

0.50 - x x x

Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)

Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50

=> Ka ≈ x^2 / 0.50

=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6

=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030

pH = - log [H+] = - log (x) = - log (0.0030) = 2.5

Answer: 2.5

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3 years ago

If you are supposed to put in 2/4 a cup sugar for a recipe, can you just put in a half cup of sugar?

natta225 [31]

I would assume so.

Given $\frac{2}{4}$ , we can simplify the fraction to $\frac{1}{2}$

Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.

Hope this is the answer you are looking for.

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3 years ago

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A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

<u>Answer:</u> The molecular formula for the compound is $C_6H_6$

<u>Explanation:</u>

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

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

Mass of C = 92.26 g

Mass of H = 7.74 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

<u>Step 2:</u> 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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

<u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$