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nordsb [41]
3 years ago
9

How many moles of Na+ ions are in 100.mL of 0.100M Na3PO4(aq)?

Chemistry
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

The answer is C) 0.0300 mol

Explanation:

First you need to get the total amount of moles is dissolved in the solution. This can be obtained doing the following:

Molarity = moles of solute / liters of solution

moles of solute = molarity x liters of solution

First, the volume has to be in liters, then :

100. ml x (1 L / 1000 mL) = 0.100 L

(We then substitute)

moles of solute = 0.100 mol/L x 0.1 L = 0.0100 mol of solute (In this case Na3PO4)

Having the moles of the solute, we now need to find how many moles of Na+ ions are there.

We need the conversion factor of 3 Na+ moles per 1 mole of Na3PO4

We then find the amount of moles doing as follows:

0.0100 mol Na3PO4 x (3 mol Na+ / 1 mol Na3PO4) = <u>0.0300 mol Na+.</u>

 

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16.05 amu

Explanation:

12.011 rounds to about 12.01 and 1.008 rounds to about 1.01 so when adding you'd do [12.01 + (1.01×4)]= 16.05

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calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation
balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= ?
  • V= 5.005 L
  • n= 1.255 mol
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K

Solving:

P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}

P= 5.62 atm

<u><em>The pressure is 5.62 atm.</em></u>

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The mass of a large order of french fries is about 170 G what is its approximate weight in pounds
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Answer:

             0.374 Pound Approximately

Solution:

Gram and Pound are related as,

1 Gram  =  0.0022 Pound (approximately)

As we are given with 170 g of French Fries, so it can be converted into pounds as,

          1 Gram  =  0.0022 Pound

So,

          170 Grams  =  X Pounds

Solving for X,

                      X =  (170 Gram × 0.0022 Pounds) ÷ 1 Gram

                      X =  0.374 Pound Approximately

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