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3 years ago

Why does magnesium nitrate not react with Magnesium sulfate?

1 answer:

3 0

Magnesium nitrate o not react with magnesium sulfate because it cannot displace the ions involved. The cations are the same so they just cancel the charges. If you look at it, even of a reaction occurs the product will still be the same substances.

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Even though lead is toxic, lead compounds were used in ancient times as white pigments in cosmetics. What is the percentage of l

Gekata [30.6K]

<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %

<u>Explanation:</u>

The given chemical formula of lead (IV) carbonate is $Pb(CO_3)_2$

To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

$\text{Mass percent of lead}=\frac{\text{Mass of lead}}{\text{Mass of lead (IV) carbonate}}\times 100$

Mass of lead = (1 × 207.2) = 207.2 g

Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g

Putting values in above equation, we get:

$\text{Mass percent of lead}=\frac{207.2g}{327.2g}\times 0100=63.32\%$

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %

4 0

2 years ago

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t

Serga [27]

Explanation:

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0

3 years ago

All atoms of the same elements have the same what?

Flura [38]

Answer:

The same number of proteins in their nucleus.

4 0

2 years ago

Read 2 more answers

How many total atoms are there in one molecule of C145H293O168?

nasty-shy [4]

Answer:

606 atoms

Explanation:

Add the numbers 145 + 293 + 168

4 0

2 years ago

g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0

2 years ago