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kramer
3 years ago
9

Given a 1.50 Liters of solution that contains 62.5 grams of magnesium oxide. What is the Molarity of that solution? Please show

all work for any credit.
Chemistry
1 answer:
zhenek [66]3 years ago
3 0
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol

Molarity: mol/L

First, find mol.

62.5 g x 1mole ÷ 40.31 g = 1.55 mol

then divide mol and the given liters

1.55mol ÷ 1.50 L= 1.03 M
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A scientist evaluates its livestock and determines which livestock it is
Rina8888 [55]

Any animals look for steangth mostly because the offspring can probobly onther trait is looks. The male will try to impess the female.

4 0
3 years ago
Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold
Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}

3 0
3 years ago
Explain how did you find the qostion
Alekssandra [29.7K]
What do you mean by this?
6 0
2 years ago
Correct answer please i need it before 12:07
choli [55]

Answer:

xenon

Explanation:

Xenon is a non-metal, odorless gas, consist of single atoms, colorless, and 8 valence electrons out of everything else on this list.

6 0
2 years ago
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