The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
Explanation:
For the given reaction:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

![Rate=k[CO]^x[H_2]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BCO%5D%5Ex%5BH_2%5D%5Ey)
where x and y are order wrt to
and 
According to collision theory , the molecules must collide for a reaction to take place. According to collision theory , the rate of a reaction is proportional to rate of collision of reactants.
Thus with an increase in concentration of reactants , the rate of reaction also increases. This is because if the concentration of reactants increases , the chances of collision between molecules also increases and thus more products wil be formed which in turn increases the rate of reaction.
Answer:
The correct answers are first, fourth, fifth
Explanation:
Answer:
1st image, 17.79 g Mg(NO3)2
2nd image 1.47 mol NH3
3rd image Molar Mass Al(OH)3 = 78 g/mol
Explanation:
1st image
Molar Mass Mg(NO3)2 = 148. 31 g/mol
0,12 mol Mg(NO3)2 x (148, 31 g Mg(NO3)2 / 1 mol Mg(NO3)2) = 17.79 gMg(NO3)2
2nd image
Molar Mass NH3 = 17g/mol
25 g NH3 x ( 1 mol NH3/ 17 g NH3) = 1.47 mol NH3
3rd image
Molar Mass Al(OH)3 = 1 Al *(27) + 3 O * (16) + 3 H * (1) = 78 g/mol