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3 years ago

Why did the water turn blue after the food coloring was added to the water

Chemistry

1 answer:

o-na [289]3 years ago

8 0

Because when mixed the food coloring spreads to the less dense parts of the water cup.

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What is the radioactive nucleus before it decays

Darina [25.2K]

Answer:

Alpha decay is one type of radioactive decay, in which an atomic nucleus emits an alpha particle, and thereby transforms (or "decays") into an atom with a mass number decreased by 4 and atomic number decreased by 2.

Explanation:

7 0

3 years ago

What kind of saccharide is this

zzz [600]

it is a disaccharide

4 0

3 years ago

How much energy is required to vaporize 185 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol

Diano4ka-milaya [45]

Answer is: 73.52 kJ of energy is required to vaporize butane.
m(C₄H₁₀) = 185 g.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 185 g ÷ 58.12 g/mol.
n(C₄H₁₀) = 3.18 mol; amount of butane.
Hvap = 23.1 kJ/mol; the heat of vaporization for butane.
Q = Hvap · n(C₄H₁₀).
Q = 23.1 kJ/mol · 3.18 mol; energy.
Q = 73.52 kJ.

4 0

4 years ago

3. What is the greatest amount of H2O that can be made with 3.8 moles of H and 5 moles of

Pachacha [2.7K]

Answer: Hello!

first i believe we need a balanced equation to start...

i got 2H2 + 1O2 = 2H2O

This tells us that we need 2 moles of H2 for every 1 mole of O2 Since we only have 1 mole of H2 compared to the 5 moles of O2 hydrogen is the limiting reagent. For illustration, divide the balanced equation by 2 in order to get 1 mole of H2 If we start with 1.0 moles of H2 we'll produce 1.0 mole of H2O

Your welcome <3

Explanation:

3 0

2 years ago

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

Where Ac⁻ is the salt of acetic acid (Acac).

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ (1)

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] (2)

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = 0,0365 mol

Ac⁻ = 0,127 M × 0,500 L = 0,0635 mol

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = 6,005 g

You need to weight 6,005 g of acetic acid

I hope it helps!

7 0

4 years ago