All categories

3 years ago

What is the mass of 3.00 moles of magnesium chloride, MgCl2? Express your answer with the appropriate units.

1 answer:

5 0

To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows: Mg=24 g/mole Cl=38 g/mole Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so: (38 g/mole*2 moles)+24 g/mole=100 g/mole Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows: 3.00 moles * 100 g/mole = 300 g

You might be interested in

No one believed him b/c he had no

Rainbow [258]

Answer:

Chicken nuggets

hope it helps have a nice day

6 0

2 years ago

Read 2 more answers

When you light a fireplace what energy transformation occurs?

snow_tiger [21]

thermal energy

hope this helped

4 0

3 years ago

The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2

SCORPION-xisa [38]

Butane is C₄H₁₀.

$C_4H_{10} + O_2 \to CO_2 + H_2O \\ \\
\hbox{balance carbon and hydrogen on the right-hand side:} \\
C_4 H_{10} + O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{balance oxygen on the left-hand side:} \\
C_4 H_{10} + \frac{13}{2} \ O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{multiply by 2 to get rid of the fraction:} \\
2 \ C_4H_{10} + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O$

The balanced equation is 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.

3 0

3 years ago

25g of potassium nitrate are dissolved in 100g of water. What is the percent by mass of potassium nitrate?

trapecia [35]

The the exact mass is 24.740% (according to the internet) but I’d say that it’s 25

8 0

3 years ago

A sample of gas with an initial volume of 12.5 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume

Kipish [7]

Answer:

Final pressure in (atm) (P1) = 6.642 atm

Explanation:

Given:

Initial volume of gas (V) = 12.5 L

Pressure (P) = 784 torr

Temperature (T) = 295 K

Final volume (V1) = 2.04 L

Final temperature (T1) = 310 K

Find:

Final pressure in (atm) (P1) = ?

Computation:

According to combine gas law method:

$\frac{PV}{T} =\frac{P1V1}{T1} \\\\\frac{(784)(12.5)}{295} =\frac{(P1)(2.04)}{310}\\\\33.22 = \frac{(P1)(2.04)}{310}\\\\P1=5,048.18877$

⇒ Final pressure (P1) = 5,048.18877 torr

⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760

⇒ Final pressure in (atm) (P1) = 6.642 atm

3 0

3 years ago