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Tanya [424]
3 years ago
14

How is heat transferred through convection?

Chemistry
2 answers:
Basile [38]3 years ago
5 0

Answer:

D. Heat is transferred by the movement of a liquid or gas.

azamat3 years ago
4 0

Answer:

The correct answer is D. I got it right on the test

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1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.
Alexxx [7]

Answer:

n=1.732mol

\% N=18.7\%

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%

Best regards!

4 0
3 years ago
What characteristics are used to clarify an area as a wetland?
Nadya [2.5K]
Soil covered, saturated, submerged, flooded w water, standing water
5 0
3 years ago
Which of the following statements does not describe what he structure of an atom an atom has a small dense center called nucleus
nikklg [1K]

Answer:

inside the nucleus of an atom are protons and electrons.

5 0
2 years ago
Please help me with this chemistry problem
Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration,  in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0
3 years ago
ODIO POD
miss Akunina [59]

Answer:

\boxed{\text{40 mol Al}}

Explanation:

            Al₂O₃ ⟶ 2Al + 3O₂

n/mol:     20

\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}

8 0
3 years ago
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