Select all that reasons that the reaction contents does not conduct at this low point.

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

1 year ago

Suppose you perform a calorimeter experiment to determine the molar heat of neutralization of an unknown acid, H A HA, with sodi

Black_prince [1.1K]

Answer:

-86.02 kJ/ mole

Explanation:

The moles of the acid used = Molarity × Volume (L) =

= 0.50 (0.0372 L)

= 0.0186 moles

The heat released = -1.6 kJ

∴ 0.0186 moles neutralization of HA heat is: -1.6 kJ

The molar heat of neutralization due to one mole of the unknown acid = -1.6/0.0186

= -86.02 kJ/ mole

5 0

2 years ago

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o

gtnhenbr [62]

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = 57.14%

5 0

2 years ago

Read 2 more answers

Click the "draw structure" button to launch the drawing utility. Propane (CH3CH2CH3) has both 1° and 2° hydrogens. Draw the carb

OlgaM077 [116]

Answer:

a) find attached image 1

b) find attached image 2

Explanation :

The more stable radical is formed by a reaction with smaller bond dissociation energy.

since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.

6 0

2 years ago

What is the name of the following ionic compound?: MoAs

Minchanka [31]

Molybdenum Arsenide

I think that’s right but not %100 sure

3 0

2 years ago