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oee [108]
3 years ago
15

A mercury thermometer is used to take four temperatures of a solution

Chemistry
1 answer:
anastassius [24]3 years ago
7 0

Answer:

scale. Mercury thermometers can be used to determine body, liquid, and vapor temperature.

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Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m
Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>;  that is

molarity = moles of solute ÷ liters of solution

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0
3 years ago
How many protons does a nitrogen atom have?<br> 07<br> 0 14<br> 8<br> 05
VLD [36.1K]

Answer:

07.

Explanation:

8 0
2 years ago
Read 2 more answers
What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


6 0
3 years ago
Did I do this science worksheet correctly? Asking for a friend
grin007 [14]

Yes everything looks good to me

8 0
3 years ago
5. Which is an example of inertia?
Hitman42 [59]

Answer:

Explanation:

D. When the truck driver slammed on the brakes, all the boxes in the back

of the truck slid forward.

6 0
3 years ago
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