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Georgia [21]
3 years ago
12

The observation that 4.0 g of hydrogen reacts with 32.0 g of oxygen to form a product with O:H mass ratio-8:1, and 6.0 g of hydr

ogen reacts with 48.0 g of oxygen to form the same product with O/H mass ratio = 8:1 is evidence for the law of 1. multiple proportions 2. erergy conservation. 3. mass conservation 4. definite proportion
Chemistry
1 answer:
irinina [24]3 years ago
8 0

Answer:

Law of definite proportion

Explanation:

As per law of definite proportion, ratio of elements present in a compound is always fixed irrespective of the  source, amount and method of preparation.

In the given case, the hydrogen and oxygen react with each other to form a compound. The ratio of oxygen and hydrogen is fixed which is 8 : 1 and this ratio does not change upon changing amount of oxygen and hydrogen.

So, the this experiment support the law of definite proportion.

Therefore, the correct option is option 4

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scZoUnD [109]

Answer:

I am not a man or boy, I am a girl so therefore I could answer this question before November ends Imao, have a great rest of your day.

Explanation:

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3 years ago
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What is true about most of the wavelengths found in the electromagnetic spectrum?
olasank [31]
D, you can't see ultraviolet rays and other similar frequency waves with the naked eye.
7 0
2 years ago
Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k
bulgar [2K]

Answer:

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

8 0
3 years ago
A concentrated aqueous solution of Pb(NO3)2 is slowly added to 1.0 L of a mixed aqueous solution containing 0.010 M Na2CrO4 and
Oduvanchick [21]

Answer:

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

Explanation:

Step 1: Data given

Molarity of Na2CrO4 = 0.010 M

Molarity of NaBr = 2.5 M

Ksp(PbCrO4) = 1.8 * 10^–14

Ksp(PbBr2) = 6.3 * 10^–6

Step 2: The balanced equation

PbCrO4 →Pb^2+ + CrO4^2-

PbBr2  → Pb^2+ + 2Br-

Step 3: Define Ksp

Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]

1.8*10^-14 = [Pb^2+] * 0.010 M

[Pb^2+] = 1.8*10^-14 /0.010

[Pb^2+] = 1.8*10^-12 M

The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M

Ksp PbBr2 = [Pb^2+][Br-]²

6.3 * 10^–6 = [Pb^2+] (2.5)²

[Pb^2+] = 1*10^-6 M

The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

5 0
3 years ago
How many grams of oxygen can be obtained from 250. grams of KCIO3?
guapka [62]

Answer:

Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.

Explanation:

3 0
2 years ago
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