Answer:
15.3 %
Explanation:
Step 1: Given data
- Mass of the sample (ms): 230 g
- Mass of carbon (mC); 136.6 g
- Mass of hydrogen (mH): 26.4 g
- Mass of nitrogen (mN): 31.8 g
Step 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
According to my astronomy book, it's about 0.206
Answer:
The final temperature is:- 7428571463.57 °C
Explanation:
The expression for the calculation of heat is shown below as:-
Where,
is the heat absorbed/released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)
Specific heat of water = 4.18 J/g°C
Initial temperature = 35 °C
Final temperature = x °C
kcal
Also, 1 kcal = 4.18 kJ =
J
So, Q =
J = 54340000 J
So,

Thus, the final temperature is:- 7428571463.57 °C
1mol=6.022x10^23 atoms
3.75mol= 6.022x3.75x10^23 atoms
=2.2583x10^24 atoms