Answer:
The original concentration is "35 mg/ml".
Explanation:
According to the question,
The solution is diluted,
= 1:50
The initial volume,
V1 = 1 ml
Final concentration,
= 0.07 mg
then,
The final volume,
V2 = 500 ml
As we know,
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
Answer:
5Atm
Explanation:
I just guess and it’s right
A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer:
The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance(mol) .
Explanation:
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