Answer:
V₂ = 1500 Liters ( 2 sig. figs.)
Explanation:
Given the following gas law variables:
P₁ = 110.0KPa P₂ = 25KPa
V₁ = 410 Liters V₂ = ?
T₁ = 17°C ( = 290K) T₂ = -27°C ( = 248K)
P₁V₁/₁T₁ = P₂V₂/T₂ => V₂ = V₁(P₁/P₂)(T₂/T₂)
V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)
V₂ = 1500 Liters ( 2 sig. figs.)
Think of it this way: it is a distribution problem in which you are multiplying the 2 on the outside with each element in the parentheses. Oxygen does not have a number in front of it, so put an imaginary one in front of it to help you. Do the same with Hydrogen since it doesn't have a number in front of it either. Now you know that hydrogen has one ion and oxygen has one... but you must now multiply each of the elements' ions by two. You should now know that Oxygen has 2 ions in Calcium Hydroxide and that there are also 2 ions of Hydrogen in Calcium Hydroxide. Does this make sense?
Answer:
Heat transferred, Q = 1542.42 J
Explanation:
Given that,
Mass of water, m = 30 grams
Initial temperature, 
Final temperature, 
We need to find the energy transferred. The energy transferred is given by :
c is specific heat of water, c = 4.18 J/g °C
So,
So, 1542.42 J of energy is transferred.
Answer:
1(a) N = 3
(b) N = 0
(c) N = 5
(d) N = -2
(2) Molecular formula for benzene is C6H6
Explanation:
1(a) N02 1-
N + (2×-2) = -1
N-4 = -1
N = -1+4 = 3
(b) N2
2(N) = 0
N = 0/2 = 0
(c) NO2Cl
N + ( 2×-2) + (-1) = 0
N - 4 - 1 = 0
N - 5 = 0
N = 0+5 = 5
(d) N2H4
2(N) + (4×1) = 0
2N + 4 = 0
2N = 0 - 4 = -4
N = -4/2 = -2
(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole
Therefore, molecular formula for benzene is C6H6
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
