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Murrr4er [49]
3 years ago
7

What is the mass of 1 mole of MgSO4?

Chemistry
1 answer:
White raven [17]3 years ago
7 0

Answer:

120.37 g

Explanation:

Mg = 24.31 g

S = 32.06 g

4 O = 16 X 4 = 64 g

Therefore, MgSO4 is (24.31 + 32.06 + 64) g = 120.37 g

You might be interested in
According to the following reaction, how many moles of chlorine gas are necessary to form 0.739 moles carbon tetrachloride? carb
Ilia_Sergeevich [38]

Answer:

2.956 moles chlorine gas will be produced

Explanation:

Step 1: data given

Number of moles carbon tetrachloride (CCl4) = 0.739 moles

carbon disulfide (s) = CS2(s)

chlorine (g) = Cl2(g)

carbon tetrachloride (l) = CCl4(l)

sulfur dichloride (s) = SCl2 (s)

Step 2: The balanced equation

CS2(s) + 4Cl2(g) → CCl4(l) +2SCl2

Step 3: Calculate moles chlorine gas

For 1 moles Cs2 we need 4 moles Cl2 to produce 1 mol CCl4 and 2 moles SCl2

For 0.739 moles CCl4 we need 4*0.739 = 2.956 moles Cl2

2.956 moles of chlorine gas will be produced

4 0
3 years ago
Which of the following is a synthesis reaction?
Allushta [10]

Answer:  4 A1+3 02 → 2 Al2O3

Explanation: Two different atoms combines together to produce a compound.

7 0
2 years ago
Read 2 more answers
5. What is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm?
Helga [31]

70.33 L  is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.

<h3>What is volume?</h3>

Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.

Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.

We will use ideal gas law to find the volume

PV = nRT

Can also be written as

V = (nRT)/P

Where,

P = pressure

V = volume

n = amount of substance

R = ideal gas constant

T = temperature

Here, we have given

P = 3.5 atm

V = to find

n = 10 moles

R = 0.08206 L⋅atm/K⋅mol

T = 300k

Lets substitute the values

V = (10 × 0.08206 × 300)/3.5

V =  70.33 L

Learn more about volume

brainly.com/question/463363

#SPJ10

3 0
1 year ago
If a student made a 2.5 solution using 1.5 moles of sucrose how many liters of water were used?
saul85 [17]

Answer:

1 mole.

Explanation:

Hello there!

2.5-1.5=1

Subtract the sucrose moles from the total moles, and there you are!

:)

3 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
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