First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na = 
.
Hence , this is the required solution .
Answer
The second part of the theory says all atoms of a given element are identical in mass and properties. The third part says compounds are combinations of two or more different types of atoms. The fourth part of the theory states that a chemical reaction is a rearrangement of atoms.
Explanation:
Answer: When the electrons in an element are more than the required, a bond is formed with other element which has deficiency of electron and in case when electrons are less in numbers the vice versa happens. Such kind of chemical bonds are known as ionic bond.
Explanation:
Answer:The new volume is 5mL
Explanation:
The formular for Boyles Law is; P1 V1 = P2 V2
Where P1 = 1st Pressure V1 = First Volume
P2 = 2nd Pressure V2 = Second Volume
From the question; P1 = 5atm, V1 = 10ml
P2 = 2 x P1 (2 x 5) = 10 atm V2 =?
Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2
∴ 5 x 10/ 10 = 5
∴ V2 = 5mL
