How many moles of Cl- ions are needed to completely combine with 0.25 moles of Mg+2 ions?

Protons are positively charged and repel other protons. Which other particle is found in the nucleus and separates protons so th

Leto [7]

Neutrons which doesn't have any charge

6 0

3 years ago

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A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach

Bogdan [553]

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ + I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

6 0

3 years ago

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the of strontium-90 is ________ yr.

lesantik [10]

The question is incomplete, here is the complete question.

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.

a. 28.8 b. 30.9 c. 35.4 d. 32.2

Answer : The half-life of strontium-90 is 28.8 years.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process = 9.00 year

a = initial amount or moles of the reactant = 1.000 g

a - x = amount or moles left after decay process = 0.805 g

Putting values in above equation, we get:

$k=\frac{2.303}{9.00}\log\frac{1.00}{0.805}$

$k=0.0241\text{ year}^{-1}$

To calculate the half-life, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$0.0241\text{ year}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=28.8\text{ years}$

Therefore, the half-life of strontium-90 is 28.8 years.

5 0

3 years ago

Which accurately labels the lysosome?<br><br> W<br> X<br> Y<br> Z

xeze [42]

I am not sure about this but I think it’s y

7 0

3 years ago

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Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

3 years ago