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tigry1 [53]
2 years ago
13

Question 10 of 25

Chemistry
1 answer:
ANEK [815]2 years ago
5 0

An example of a case in which a wave is absorbed by matter is light seen through a glass window.

<h3>Can matter absorb a wave?</h3>

Matter is district from waves. A wave is a form of energy which could be absorbed by matter on its path. This is common with both mechanical and electromagnetic waves.

Hence, an example of a case in which a wave is absorbed by matter is light seen through a glass window.

Learn more about matter and waves:brainly.com/question/21180536

#SPJ1

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You want to determine the protein content in milk with the Kjeldahl method. You take 100 g whole milk and use 100 mL of 0.5 M hy
stepladder [879]

Answer:

3.38%

Explanation:

Given that;

the mass of the whole milk sample = 100 g

volume of HCl = 100 mL = 0.1 L

molarity of HCl = 0.5 M

volume of NaOH = 34.50 mL = 0.0345 L

molarity of NaOH = 0.3512 M

Since we knew the molarity and volume of both HCl and NaOH; we can calculate their corresponding number of moles present.

So, number of moles of HCl = molarity of HCl × volume of HCl

number of moles of HCl = 0.5 M ×  0.100 mL

                                        = 0.05 mole

number of moles of NaOH = Molarity of NaOH ×  Volume of NaOH

number of moles of NaOH = 0.3512 M ×  0.0345 L

                                            = 0.012 mole

From the question, we can deduce that the number of HCl that is consumed by NH₃ is equal to the number of moles of HCl that is consumed by NaOH.

SO, number of moles of HCl consumed by NH₃ = Total moles of HCl - moles of HCl consumed by NaOH

= 0.05 mole - 0.012 mole

= 0.038 mole

However, to determine the mole of NH₃ present , we have:

number of moles of NH₃ present = number of moles of HCl consumed by NH₃  = 0.038

∴ the mass of Nitrogen with the molecular weight (14.0 g/mol) = 0.038 moles × 14.0 g/mol

= 0.530 g

Now, the percentage of Nitrogen can be calculated as;

percentage of nitrogen =\frac{mass of nitrogen}{mass of the whole milk sample} *100

percentage of nitrogen =\frac{0.530g}{100g} *100

percentage of nitrogen =0.530%%

the percentage of protein in the sample = CP × %age of N

where CP is given as 6.38

∴ the percentage of protein in the sample = 6.38 ×  0.530%

the percentage of protein in the sample = 3.3814%

the percentage of protein in the sample = 3.38%

6 0
3 years ago
What is the empirical formula of CoH1803? (4 points)
butalik [34]

Answer:

-O CH30

Explanation:

My Head Is About To Explode

7 0
3 years ago
The Sun is a yellow star that's both average in brightness and temperature and is classified as
Roman55 [17]

Answer:

Classifying stars according to their spectrum is a very powerful way to begin to understand how they work.  As we said last time, the spectral sequence O, B, A, F, G, K, M is a temperature sequence, with the hottest stars being of type O (surface temperatures 30,000-40,000 K), and the coolest stars being of type M (surface temperatures around 3,000 K).  Because hot stars are blue, and cool stars are red, the temperature sequence is also a color sequence.  It is sometimes helpful, though, to classify objects according to two different properties.  Let's say we try to classify stars according to their apparent brightness, also.  We could make a plot with color on one axis, and apparent brightness on the other axis, like this:

Explanation:

7 0
3 years ago
Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

4 0
3 years ago
Which of the following bonds is the strongest?
mr Goodwill [35]
The answer to this question will be c
3 0
3 years ago
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