All categories

3 years ago

Determine the number of grams of magnesium oxide if you have 2.63 moles of magnesium oxide.

Chemistry

1 answer:

ANTONII [103]3 years ago

3 0

The number of grams of Magnesium oxide= 106 g

<h3>Further explanation</h3>

Given

2.63 moles of Magnesium oxide

Required

mass of Magnesium oxide

Solution

The mole is the number of particles contained in a substance

Moles can also be determined from the amount of substance mass and its molar mass

Mol = mass : Molar mass

Mass of Magnesium oxide(MgO, MW=40.304 g/mol) :

= mol x MW

= 2.63 x 40.304

= 106 g

You might be interested in

What would be the effect of bringing a negatively charged metal ball near an iron bar

tensa zangetsu [6.8K]

Because The bar would become negatively charged and repel the metal ball. A magnetic north pole would be generated on the bar, and it would repel the metal ball.

7 0

3 years ago

Which of the following molecules is nonpolar?

Dvinal [7]

Answer:

CO2

Explanation:

There are two types of molecules

Polar
Non polar

Non polar molecules are insoluble in water .

7 0

3 years ago

What are some ways we can improve the chemical reactions in batteries to increase their efficiency?

Ainat [17]

Answer:

Charging Level The cycle life of Lithium batteries can be increased by reducing the charging cut off voltage.

Explanation:

6 0

3 years ago

Bleach is more basic than egg whites.

olga nikolaevna [1]

Eggs whites are more acidic than bleach

4 0

3 years ago

A gas has an initial volume of 52.3 L at 273 Kelvin. What is its temperature when the volume reaches 145.7 L?

erastovalidia [21]

Answer:

$\boxed {\boxed {\sf 761 \ K}}$

Explanation:

We are asked to find the new temperature of a gas after a change in volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume is initially 52.3 liters at a temperature of 273 Kelvin.

$\frac {52.3 \ L}{273 \ K}= \frac{V_2}{T_2}$

The volume reaches 145.7 liters at an unknown temperature.

$\frac {52.3 \ L}{273 \ K}= \frac{145.7 \ L }{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$52.3 \ L * T_2 = 273 \ K * 145.7 \ L$

Now the variable is being multiplied by 52.3 liters. The inverse of multiplication is division. Divide both sides by 52.3 L.

$\frac {52.3 \ L * T_2 }{52.3 \ L}=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

$T_2=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

The units of liters cancel.

$T_2=\frac{ 273 \ K * 145.7 }{52.3 }$

$T_2 = 760.5372849 \ K$

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we found, that is the ones place. The 5 in the tenths place tells us to round the 0 up to a 1.

$T_2 \approx 761 \ K$

When the volume reaches 145.7 liters, the temperature is <u>761 Kelvin.</u>

8 0

3 years ago