Answer:The molecular formula of the oxide of metal be . The balanced equation for the reaction is given by:
Explanation:
Let the molecular formula of the oxide of metal be
Mass of metal product = 1.68 g
Moles of metal X =
1 mol of metal oxide produces 2 moles of metal X.
Then 0.03005 moles of metal X will be produced by:
of metal oxide
Mass of 0.01502 mol of metal oxide = 2.40 g (given)
y = 2.999 ≈ 3
The molecular formula of the oxide of metal be . The balanced equation for the reaction is given by:
Answer:
The correct answer is;
Demand for gasoline in Orlando is price inelastic.
Explanation:
The elasticity is the degree of response to a change in price or quantity supplied to the the quantity demanded. An elastic demand responds positively to change in price, while an inelastic demand means that when there is a price increase, the quantity demanded remains the same and where there is a drop in price the quantity demanded remains constant.
If a small change in price results in a large change in demand then the good is said to be price elastic
In the question the price increases by 10% while the quantity demanded drops 5 % daily. Therefore it is price inelastic
Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O