All categories

1 year ago

If we compare a mole of carbon atoms to a mole of oxygen atoms, which sample will contain more atoms?.

1 answer:

3 0

Since 1 mole of any element contains 6.0221023 atoms, a mole of carbon and a mole of oxygen will have the same number of atoms.

<h3>Is a mole of oxygen equivalent to a mole of carbon?</h3>

Answer and justification The molar mass of carbon differs from the molar mass of oxygen, hence this assertion is untrue. As a result, one mole of oxygen and one mole of carbon cannot have the same mass.

<h3>What kind of link unites carbon and oxygen?</h3>

A polar covalent link between carbon and oxygen is known as a carbon-oxygen bond. With six valence electrons, oxygen likes to either share two of them in a bond with carbon, leaving the remaining four unoccupied.

To know more about mole of carbon atoms visit:-

brainly.com/question/20484279

#SPJ4

You might be interested in

Rpd pls ii test :)<br><br> tot ce se afla in poza

mamaluj [8]

Answer:

Sorry mate! I can't understand this language...

4 0

2 years ago

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

5 0

2 years ago

A gas occupies a volume of 31.0 ml at 19.0°C. If the gas temperature rises to 38.0°C at constant pressure. Calculate the new vol

Alik [6]

Answer:

The correct answer is 0.0033 L (33.0 mL)

Explanation:

We uses the Charles's law which describes the changes in the volume (V) of a gas and its temperature in Kelvin (T) at constant pressure. The mathematical expression is the following:

V₁/T₁ = V₂/T₂

We have the following data:

V₁= 31.0 mL = 0.0031 L

T₁= 19.0°C = 292 K

T₂= 38.0°C = 311 K

V₂= ?

We calculate V₂ from the mathematical expression, as follows:

V₂= V₁/T₁ x T₂ = 0.0031 L/(292 K) x 311 K = 0.0033 L

5 0

2 years ago

Give the hybridization for the O in H3O+

Damm [24]

Answer: It would be Sp2, because H3O+ has planar structure.

4 0

3 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

3 years ago