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mamaluj [8]
2 years ago
6

Lets say you were doing an experiment on two different chemicals in two different test tubes. One chemical is giving off oxygen

gas or hydrogen gas the other is giving off carbon dioxide gas. What one test could you do in each test tube that would prove which is giving of the oxygen or hydrogen an which is giving off the carbon dioxide
Chemistry
1 answer:
boyakko [2]2 years ago
5 0

Answer:

Tests for gases

Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.

Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.

Oxygen. A glowing wooden splint relights in a test tube of oxygen.

hope it will help

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diagram 1 above shows equimolar samples of two gases inside a container fitted with a removable barrier placed so that each gas
vampirchik [111]

This problem is describing the state two gases have when separated and together as shown on the attached picture. First of all, diagram 1 shows how they are separated in two containers with apparently equal volumes, whereas diagram 2 shows the removal of the barrier so that they get mixed together.

In this case, we can analyze that each gas has its own pressure and due to the removal of the barrier, both pressure and volume undergo a change. Thus, we can infer that the final volume is doubled with respected to the initial one for each gas, causing the pressure of each gas to be halved and the total pressure the half of the added ones, in agreement to the Boyle's law (inversely proportional relationship between pressure and temperature).

Therefore, the correct choice is:

C. The partial pressure of each gas in the mixture is half its initial pressure; the final total pressure is half the sum of the initial pressures of the two gases.

Learn more:

  • brainly.com/question/21184611

8 0
2 years ago
What two types of matter are closest in<br> density? How do you know?<br> ings.
maks197457 [2]

Answer:

Density: The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance.

In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them

I hope it helps you

7 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
What are the basic properties of waves
IRINA_888 [86]

Answer:

There are many properties that scientists use to describe waves. They include amplitude, frequency, period, wavelength, speed, and phase. Each of these properties is described in more detail below. When drawing a wave or looking at a wave on a graph, we draw the wave as a snapshot in time.

Explanation:

3 0
3 years ago
Read 2 more answers
I don't know how to do this can I get the answers plz it's due in 1 hour
katrin [286]
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
4 0
3 years ago
Read 2 more answers
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