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3 years ago

Which of the following compounds will not produce O2 gas when it decomposes? (2 points)

Chemistry

1 answer:

Umnica [9.8K]3 years ago

6 0

LiOH is the right answer.

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Which is a characteristics of all ions

Sholpan [36]

Answer:

All ions must a overall charge

Explanation:

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3 years ago

What is the correct name for the n3– ion? nitride ion nitrogen ion nitrite ion nitrate ion

dem82 [27]

The correct name for the N3- ion is a nitride ion.

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3 years ago

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Which statement accurately describes radioactive dating?

geniusboy [140]

Answer:

its d and can i get brainliest plz:)

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2 years ago

How many grams of phosphorus are in 500.0 grams of calcium phosphide? (i need the work also)

kvv77 [185]

Answer:

$\boxed{\text{170.0 g P}}$

Explanation:

The formula of calcium nitride is Ca₃P₂.

The masses of each element are:

$\begin{array}{lrcr}\text{3Ca:} & 3 \times 40.08&=& \text{120.24 u}\\\text{2P:} & 2\times 30.97&=& \text{61.94 u}\\& \text{TOTAL} & = & \text{182.18 u}\\\end{array}$

So, there are 61.94 g of P in 182.18 g of Ca₃P₂.

In 500 g of Ca₃P₂:

$\text{Mass of P} = \text{500.0 g Ca$_{3}$P$_{2}$} \times \dfrac{\text{61.94 g P}}{\text{182.18 g Ca$_{3}$P$_{2}$}} = \text{170.0 g P}$

There are $\boxed{\textbf{170.0 g P}}$ in 500.0 g of Ca₃P₂.

3 0

3 years ago

A gas mixture with a total pressure of 750 mmHg contains each of the following gases at the indicated partial pressures: CO2 , 1

jolli1 [7]

Answer: a) 211 mm Hg

b) 0.629 grams

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2+p_3+p_4$

$p_{total}$ = total pressure = 750 mmHg

$p_{CO_2}$ = 124 mm Hg

$p_{Ar}$ = 218 mm Hg

$p_{O_2}$ = 197 mm Hg

$p_{He}$ = ?

$750 mmHg=124 mm Hg+218 mm Hg+197 mm Hg+p_{He}$

$p_{He}=211mmHg$

Thus the partial pressure of the helium gas is 211 mmHg.

b) According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 211 mmHg = 0.28 atm (760mmHg=1atm)

V= Volume of the gas = 13.0 L

T= Temperature of the gas = 282 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

$n=\frac{PV}{RT}=\frac{0.28\times 13.0}0.0821\times 282}=0.157moles$

Mass of helium= $moles\times {\text {molar mass}}=0.157\times 4=0.629g$

Thus mass of helium gas present in a 13.0-L sample of this mixture at 282 K is 0.629 grams

8 0

3 years ago