Describe a procedure to separate a mixture of sugar black pepper and pebbles

of the following warnings, which one refers to a chemical property of a substance? fragile flammable handle with care shake well

coldgirl [10]

Flammable is ur answer

6 0

3 years ago

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

Answer: The ratio of carbon in both the compounds is 1 : 2

Explanation:

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

For Sample 1:

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

For Sample 2:

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

3 years ago

What needs to be considered in balancing redox reactions?

Degger [83]

Number D both H and OH ions

8 0

3 years ago

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade

egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

Explanation:

Assuming Pressure is constant.

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

$\frac{300}{273} = \frac{V_2}{288}$

$V_2 = \frac{288*300}{273} \\V_2 = 316.5$

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

5 0

2 years ago

Resistance depends on which properties of a wire

Paraphin [41]

Resistance depends on there properties of a wire:
- length (the longer the more resistance)
- area (the less area = more resistance and vise verse more area = less resistance)
- resistivity (the more resistivity the resistance)

Hope this helps :)

7 0

3 years ago

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