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3 years ago

All of these properties are for alkali metals except one

1 answer:

4 0

The answer is high melting and boiling points. Because alkali metals have a range of melting points from a high of 179 °C (354 °F) for lithium to a low of 28.5 °C (83.3 °F) for cesium

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So if there is 16 protons and 16 neutrons what would be the charge right now?

sveta [45]

Answer:

Explanation:

Protons have a positive net charge,

Neutrons being neutral don't have a net charge.

Each proton is one extra net charge if you have 16 of them and the neutrons don't affect your net charge you will have 16.

6 0

3 years ago

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

$\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}$

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

$Density=\frac{Mass}{Volume}$

Now putting all the given values in this formula, we get:

$3.52g/cm^3=\frac{284.4g}{Volume}$

Volume = $80.8cm^3$

As we know that:

$1cm^3=0.061inch^3$

So,

Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

Therefore, the volume of pure diamond is $0.493inch^3$

5 0

3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$