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3 years ago

An uncharged atom of boron has an atomic number of 5 and an atomic mass of 11. how many protons does boron have?

Chemistry

2 answers:

Ede4ka [16]3 years ago

7 0

Atomic number is equal to protons and electrons, so the answer is 5

defon3 years ago

5 0

It has 5 protons. Usually if an atom is neutral, the atomic number and the number of protons are the same. <span />

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A weather map shows red semicircles on a line. What type of weather would one expect?

gtnhenbr [62]

A warm front brings drizzly weather.

3 0

3 years ago

Read 2 more answers

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers

Calculate the percentage by mass of the indicated element in the following compounds. Hydrogen in ascorbic acid, HC6H7O6, also k

Scorpion4ik [409]

Answer: The mass percent of hydrogen in ascorbic acid is 4.5 %

Explanation:

In $C_6H_8O_6$ , there are 6 carbon atoms, 8 hydrogen atoms and 6 oxygen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

$\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Molar mass of ascorbic acid}}\times 100$

Mass of hydrogen = $8\times 1g/mol=8g$

Molar Mass of ascorbic acid = $6\times 12g/mol+8\times 1g/mol+6\times 16g/mol=176g$

Putting values in above equation, we get:

$\text{Mass percent of hydrogen}=\frac{8g}{176}\times 100=4.5\%$

Hence, the mass percent of of hydrogen in ascorbic acid is 4.5 %.

8 0

3 years ago

The solution you identified in question (1) acts as a buffer due to reactions that occur within the solution when an acid or a b

weeeeeb [17]

Answer:

The answer is " $\bold{CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)}$ "

Explanation:

When $HCI$ is added in the chemical equation it reacts with sodium acetate so, it will give the following chemical equation:

$CH_3COONa\ (aq) + HCl\ (aq)\longrightarrow CH_3COOH\ (aq) + NaCl\ (aq)\\\\$

In this, the $CH_3COOH$ is a weak acid so, it not completely dissociated.

$CH_3COONa \ (aq) \ \ and \ NaCl$ were strong electrolytes they are completely dissociated.

The $HCl$ is a strong acid so, it is completely dissociated So, the net ionic equation is:

$CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)$

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3 years ago

What is the mass of 18.0 mL of honey if its' density is 1.42 g/mL?

BigorU [14]

mass (m) = ? , volume = 18.0 ml , density = 1.42 g/ml .

$density = \frac{mass}{volume} \\ \\ d = \frac{m}{v} \\ \\ mass = density \times volume \\ \\ m = d \times v \\ \\ m = 18 \: ml \times \: 1.42 \: \frac{g}{ml} \\ \\ m = 25.56 \: g$

I hope I helped you^_^

8 0

3 years ago