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3 years ago

What is the redox equation for sulphuric acid and magnesium ribbon

1 answer:

8 0

Sulfuric acid: $H_2SO_4_{(aq)}$
Magnesium ribbon: $Mg_(s)$
$Mg_{(s)}+H_2SO_4_{(aq)} \rightarrow MgSO_4_{(aq)}+H_2_{(g)}$

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1. How is the genome like a stack of papers?

Ivanshal [37]

Answer:

Except for mature red blood cells, all human cells contain a complete genome. DNA in the human genome is arranged into 24 distinct chromosomes—physically separate molecules that range in length from about 50 mil- lion to 250 million base pairs.

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2 years ago

Read 2 more answers

Avery has a glass of pure water. She pours a tablespoon of sugar into the water to make it taste sweet.

SSSSS [86.1K]

The first blank is a compound, second is a mixture

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4 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

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3 years ago

Corn plants may be tall (TT or Tt) or short (tt). The breeding of two parent corn plants results in offspring that are all tall.

zubka84 [21]

Answer:

Explanation:

The probability that both of the parents are homozygous recessive is zero.
In our case, The gene for tallness is dominant over the gene for dwarfism
Therefore; TT is homozygous dominant while tt is homozygous recessive and Tt is heterozygous.
Heterozygous genotype (Tt) will exhibit a tall phenotype.
When two parents are crossed and all the offsprings are tall then the possible genotype of the parents is either;

Homozygous dominant for both parents or

TT x TT

2. One parent is homozygous dominant and the other is homozygous recessive.

TT x Tt

Therefore, the probability of having both parents as homozygous recessive when all the offspring are tall is Zero.

8 0

3 years ago

Hello :) for qn 3 , I got 0.01 mol but I’m not sure if it’s correct :/ I need help , thanks!

damaskus [11]

Answer:

0.05 dm³

Explanation:

Please see the attached picture for full solution.

The question is asking for the volume of H₂SO₄, so we need to find the number of moles of KOH then the number of moles of H₂SO₄. (Using mole ratio from the balanced equation) Also, potassium hydroxide is KOH not K₂SO₄ :)

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3 years ago