Meiosis 1 is the reductional division ! so after telophase 1 , cytokinesis takes place and two haploid cells are formed !
Eg: 2H2 + O2 ➡ 2H2O
This shows that sum of reproducts = sum of products.
Therefore, mass is conserved.
I hope this helps you.
Answer:
The answer to your question is 29.4 grams of H₂SO₄
Explanation:
Data
mass of sulfuric acid
volume = 50 ml
Molarity = 6 M
Process
1.- Write the formula of Molarity
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = 6 x 0.05
-result
moles = 0.3
2.- Calculate the molar mass of Sulfuric acid
H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 2 + 32 + 64 = 98 g
3.- Calculate the mass in the solution
98 g of H₂SO₄ ---------------- 1 mol
x ----------------- 0.3 moles
x = (0.3 x 98) / 1
x = 29.4 grams of H₂SO₄
Answer:
The correct option is;
B) 179 g
Explanation:
The parameters given are;
Mass of H₂ that takes part in the reaction = 2.23 g
Molar mass of hydrogen gas, H₂ = 2.016 g
Number of moles, n, of hydrogen gas H₂ is given by the relation;
Chemical equation for the reaction;
H₂ + Br₂ → 2HBr
Given that one mole of H₂ reacts with one mole of Br₂ to produce two moles of HBr
1.106 mole of H₂ will react with 1.106 mole of Br₂ to produce 2 × 1.106 which is 2.212 moles of HBr
The molar mass, of HBr = 80.91 g/mol
The mass of HBr produced = Molar mass of HBr × Number of moles of HBr
The mass of HBr produced = 80.91 × 2.212 = 178.997 g ≈ 179 grams
Therefore, the correct option is B) 179 g.
From the reduction standard potentials;
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
