Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
It is codominant inheritance because, if the placement of the A and B molecules on each cell is controlled by the proteins that are coded by different versions of the same gene, then <span>IA and IB </span><span>are codominant but both are dominant to I<span>o</span>. If a person receives an <span>IA </span>allele and a <span>IB</span> allele, their blood type is type AB, in which characteristics of both A and B antigens are expressed.
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Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
Is there answer choices ??
The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>
Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
