Answer:
0.24M
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation above, we obtained the following information:
nA (mole of acid) = 1
nB (mole of base) = 2
Data obtained from the question include:
Va (volume of the acid) = 12mL
Ca (concentration of the acid) =?
Vb (volume of the base) = 36mL
Cb (concentration of the base) = 0.16 M
The Ca (concentration of the acid) can be obtained as follow:
CaVa/CbVb = nA/nB
Ca x 12 / 0.16 x 36 = 1 /2
Cross multiply to express in linear form as shown below:
Ca x 12 x 2 = 0.16 x 36
Divide both side by 12 x 2
Ca = 0.16 x 36/ 12 x 2
Ca = 0.24M
Therefore, the concentration of the acid is 0.24M
Basically since there’s 2 hydrogen’s there will be two H’s on either side of your other element. And Se is in group 6 which means it has 6 valence electrons. When you combine 6 and 2 from the hydrogen you get 8. You then should place 8 dots around Se two on each side.
So something like H Se(with 8 dogs around it) H
From the coefficients of the equation, we know that for every 3 moles of water consumed, 1 mole of diphosphorus trioxide is consumed.
This means we need to find the mass of 0.75 moles of diphosphorus trioxide.
- The atomic mass of phosphorous is 30.973761998 g/mol.
- The atomic mass of oxygen is 15.9994 g/mol.
So, the formula mass of diphosphorus trioxide is:
- 2(30.973761998)+3(15.9994)=109.945723996 g/mol.
Thus, 0.75 moles have a mass of:
- 0.75(109.945723996), which is about 82.5 g (to 3 sf)
Answer:
6.25 g of Carbon is in Excess
Explanation:
The balance chemical equation is as follow:
C + O₂ → CO₂
Step 1: Calculate Moles of O₂ as:
Moles = Mass / M/Mass
Moles = 50 g / 32 g/mol
Moles = 1.562 mol of O₂
Step 2: Calculate Moles of C as;
Moles = 25 g / 12 g/mol
Moles = 2.08 mol
Step 3: Calculate the Limiting reagent:
According to equation,
1 moles of O₂ reacted with = 1 mole of C
So,
1.562 moles of O₂ will react with = X moles of C
Solving for X,
X = 1 mol × 1.562 mol / 1 mol
X = 1.562 mol of C
While we are provided with 2.08 mol of C. Means C is in excess.
Step 4: Calculate amount of Excess C as;
Excess Moles = Given Moles - Used Moles
Excess Moles = 2.08 mol - 1.562 mol
Excess Moles = 0.521 moles
Step 5: Converting Excess moles to Mass as:
Mass = Moles × M.Mass
Mass = 0.521 mol × 12 g/mol
Mass = 6.25 g
