Heat, gas production, and color change can all be evidence of chemical change.

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

If the atoms in an ionic bond are not sharing electrons, what keeps the atoms together?

faltersainse [42]

Answer:

Oppositely charged particles attract each other. This attractive force is often referred to as an electrostatic force. An ionic bond is the electrostatic force that holds ions together in an ionic compound.

Explanation:

Hope this helps :)

6 0

3 years ago

Which characteristics best identify differences between intrusive and extrusive

tresset_1 [31]

Intrusive -inside
extrusive -outside

8 0

3 years ago

A sample that is 74.5 percent chloride by mass is dissolved in water and treated with an excess of

12345 [234]

Answer:

Total mass of original sample = 0.38 g

Explanation:

Percentage of chloride = 74.5%

Mass of AgCl precipitate = 1.115 g

Mass of original sample = ?

Solution:

Mass of chloride:

1.115 g/ 143.3 g/mol × 35.5 g/mol

0.0078 g × 35.5 = 0.28 g

Total mass of compound:

Total mass = mass of chloride / %of Cl × 100%

Total mass = 0.28 g/ 74.5% × 100%

Total mass = 0.0038 g× 100

Total mass = 0.38 g

5 0

3 years ago

Carbon and other non-metals are found in which area of the periodic table?

Luden [163]

Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq). The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2).

on the left-most side - alkali metals
on the right side - Non-Metals
in the middle column of the periodic table - transition metals
in the bottom row - lanthanides and the actinides

3 0

3 years ago