Answer:
Volume V2 = 948.13 ml
Explanation:
Given:
Volume V1 = 525 ml
Temperature T1 = -25°C + 273.15
Volume V2 = ?
Temperature T1 = 175°C + 273.15
Computation:
V1 / T1 = V2 / T2
525 / [-25°C + 273.15] = V2 / [175°C + 273.15]
Volume V2 = 948.13 ml
Answer:
Molecular formula = C₅H₁₀
Explanation:
From the question given above, the following data were obtained:
Empirical formula of compound => CH₂
Molar mass of compound = 70.1 amu
Molecular formula of compound =...?
The molecular formula of a compound is usually a multiple (n) of the empirical formula i.e
Molecular formula = [CH₂]ₙ
Thus, to obtain the molecular formula of the compound, we must first determine the value of n. The value of n can be obtained as follow:
[CH₂]ₙ = 70.1
[12 + (2×1)]n = 70.1
[12 + 2]n = 70.1
14n = 70.1
Divide both side by 14
n = 70.1 / 14
n = 5
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₅
Molecular formula = C₅H₁₀
Thus, the Molecular formula of the compound is C₅H₁₀.
Explanation:
HOPE THIS HELPS!
<u>Answer:</u> The number of phosphorus atoms in given amount of copper(II) phosphate is
<u>Explanation:</u>
We are given:
Moles of copper(II) phosphate = 7.00 mol
1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atoms
Moles of phosphorus in copper(II) phosphate =
According to the mole concept:
1 mole of a compound contains number of particles
So, 7.00 moles of copper(II) phosphate will contain = number of phosphorus atoms.
Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate =
Moles of sodium dichromate =
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.