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Anastasy [175]
2 years ago
15

How many moles are in 2.00 g of NaCl?7

Chemistry
1 answer:
coldgirl [10]2 years ago
7 0

Answer:

              0.0342 moles

Explanation:

Moles and Mass are related as,

Moles = Mass / M.Mass

M.Mass of NaCl = 58.44 g/mol

Putting values,

Moles =  2.00 g / 58.44 g/mol

Moles = 0.0342 moles

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Help pls! thank u (:
alexgriva [62]

Answer:

7.65 moles of silver are produced

Explanation:

Zinc, Zn, reacts with silver nitrate, AgNO3, as follows:

Zn + 2AgNO3 → Zn(NO3)2 + 2Ag

<em>Where 1 mole of Zn reacts with an excess of AgNO3 to produce 2 moles of Ag</em>

To solve this question we must convert the mass of Zn to moles and, using the chemical equation, we can find the moles of Ag as follows:

<em>Moles Zn (Molar mass: 65.38g/mol):</em>

250g Zn * (1mol / 65.38g) = 3.824 moles Zn

<em>Moles Ag:</em>

3.824 moles Zn * (2mol Ag / 1mol Zn) =

<h3>7.65 moles of silver are produced</h3>
7 0
2 years ago
Cl2 + AlBr3 → AlCl3 + Br2How many grams of Cl2 are needed to produce 19.03g of AlCl3?
VladimirAG [237]

Answer:

14.89 g

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 0
3 years ago
Jason ran 5/7 of the distance around the school track sara ran 4/5 of Jason distance what fraction of the total distance around
FrozenT [24]
Let x represent the total distance around the track
Jason's distance: (5/7)x
Sara ran (4/5) of Jason's distance,
so she ran (4/5)*(5/7)x = (4/7)x
Sara ran 4/7 of the total distance
7 0
3 years ago
A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l
sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

PV=nRT

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between  pressure and number of moles of gas will be:

\frac{P_1}{P_2}=\frac{n_1}{n_2}

where,

P_1 = initial pressure of gas = 24.5 atm

P_2 = final pressure of gas = 5.30 atm

n_1 = initial number of moles of gas = 1.40 moles

n_2 = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}

n_2=0.301mol

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0
2 years ago
Heating galactose, a monosaccharide sugar, in the presence of excess oxygen produces carbon dioxide gas and water vapor. Classif
artcher [175]

Answer:

it's a combustion reaction

6 0
2 years ago
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