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3 years ago

How many grams of NaCI are present in 11.00 moles

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

7 0

Answer: 642.93 g of NaCl

Explanation:

11.00 mol NaCl x __58.448 g__ = 642.928 g of NaCl

1 mol NaCl

I would round to 642.93 g of NaCl, but round to however many significant figures asked for.

My chemistry teacher made this map (image attached) to teach us how to do conversions. Just follow the map. I think it's pretty straight forward. I hope this helps.

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At the start of a measurement, a radioisotope has 10,000 unstable nuclei. over 2 days 5,000 of these unstable nuclei undergo rad

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The half life of a radioactive element is the time needed to the element to decay and reach the half amount of the initial amount. Here we have a radioisotope element which decays its half from 10,000 to 5,000 in two days. Therefore, its half life is 2 days.

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3 years ago

What voltage will produce 3 A through a circuit that has 15 ohms of resistance?

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Answer:

5 ohms of resistance

Explanation:

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3 years ago

Read 2 more answers

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

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2 years ago

Which two species are most closely related

Eddi Din [679]

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3 years ago

One mole of an ideal gas occupies a volume of 22.4 liters at

-Dominant- [34]

Answer : One mole of an ideal gas occupies a volume of 22.4 liters at STP.

Explanation :

As we know that 1 mole of substance occupies 22.4 liter volume of gas at STP conditions.

STP stands for standard temperature and pressure condition.

At STP, pressure is 1 atm and temperature is 273 K.

By using STP conditions, we get the volume of 22.47 liter.

Hnece, the one mole of an ideal gas occupies a volume of 22.4 liters at STP.

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3 years ago