Atom can never be divided into smaller particles is the answer was disproved
The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
Active transport is the moving of molecules across the membrane of the cell against the concentration gradient with the use of ATP.
Low to high concentration. Concentration gradient is the diffusion (movement of molecules from regions of low concentration) from high to low with the gradient. Active transport is from low to high, against the gradient.
