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mariarad [96]
3 years ago
6

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃ (My subject is not really chemistry I had no other option)

Chemistry
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

hi      it is  16c

Explanation:

hi

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3 examples of density synonyms
DENIUS [597]

frequency.

quantity.

thickness.

body.

closeness.

concretion.

heaviness.

solidity.

3 0
3 years ago
Convert 14.72 kg to ____ mg
navik [9.2K]

Answer:

14720000

Explanation:

1 kg = 1000000 mg

14.72 kg = 14.72 x 1000000

=14720000

Please Mark me brainliest

6 0
2 years ago
0.25 mol of nitric acid is used to make a 0.10 M nitric acid solution. How many grams of solute was used?
Yanka [14]

Answer:

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

Explanation:

Step 1: Data given

Nitric acid = HNO3

Molar mass of H = 1.01 g/mol

Molar mass of N = 14.0 g/mol

Molar mass O = 16.0 g/mol

Number of moles nitric acid (HNO3) = 0.25 moles

Molairty = 0.10 M

Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

Molar mass HNO3 = 1.01 + 14.0 + 3*16.0

Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.25 moles * 63.01 g/mol

Mass HNO3 = 15.75 grams

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

7 0
3 years ago
Are these correct? Thank you!! If they are not, which do I fix?
Gennadij [26K]
This looks correct to me! <3 have a good day
3 0
2 years ago
Read 2 more answers
Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

4 0
3 years ago
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