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3 years ago

Which of the following polyatomic ions will form an ionic compound with two sodium ions? CO32− HCO31− NO21− NO31−

Chemistry

1 answer:

BaLLatris [955]3 years ago

8 0

Answer:

CO32−

Explanation:

We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.

When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;

2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)

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What mass of O2 is needed to produce 4.5 moles of CO2?

Ahat [919]

Answer:

Explanation:

4 0

3 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

2 years ago

A solution of 0.0027 M K2CrO4 was diluted from 3.00 mL to 100. mL. What is the molarity of the new solution?

Stells [14]

Answer:

[K₂CrO₄] → 8.1×10⁻⁵ M

Explanation:

First of all, you may know that if you dilute, molarity must decrease.

In the first solution we need to calculate the mmoles:

M = mmol/mL

mL . M = mmol

0.0027 mmol/mL . 3mL = 0.0081 mmoles

These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.

New molarity is:

0.0081 mmoles / 100mL = 8.1×10⁻⁵ M

4 0

2 years ago

Answer this i boot you

noname [10]

Answer: answer what

Explanation:

7 0

2 years ago

If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro

In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.

Molality=

$\frac{Number of moles of solute}{Mass of solvent in kg}$

Number of moles of solute, n=

$\frac{Given mass of the substance}{Molar mass of the substance}$

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,

$n= \frac{0.2 }{123.06}$

Number of moles of nitrobenzene, n= 0.0016 mol

Mass of 10.9 g of naphthalene in kg=0.0109

$Molality= \frac{0.0016}{0.0109 }$

Molality= 0.146 m

7 0

3 years ago