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2 years ago

How do most diamonds get their color?

1 answer:

5 0

Answer:Most naturally-coloured diamonds are created when trace elements interact with the carbon atoms during the diamond's creation. The presence of chemical elements such as nitrogen, sulphur, and boron can colour diamonds in shades of yellow, green, and blue. ... Trace elements have never been found in pink diamonds.

in other words it depends on the heat of how it is made

You might be interested in

How much pure acid is in 520 milliliters of a 13 % solution?

Liono4ka [1.6K]

It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.

Volume of pure acid = 520*0.13 = 67.6 mL

4 0

2 years ago

If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?

Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

Volume of Hexane dispensed = 40ml

Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

brainly.com/question/4721701

7 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

A reaction is non-spontaneous at any temperature when

slavikrds [6]

Answer:

A reaction is non-spontaneous at any temperature when the Gibbs free energy > 0.

Explanation:

There is a state function, that determines if a reaction is sponaneous or non spontaneous:

ΔG = Gibbs free energy

A reaction is non spontaneous when it does require energy to produce that reaction. It will be spontaneous, when the reaction does not require energy to be occured.

The formula is: ΔG = ΔH - T.ΔS

ΔH → Enthalpy → Energy gained or realeased as heat.

ΔH < 0 → <em>Exothermic reaction. Spontaneity is favored </em>

T → Temperature

ΔS → Entropy → Degree of disorder of a system.

When the system has a considered disorder ΔS > 0, disorder increases.

When the system is more ordered, ΔS < 0, disorder decreases.

The reaction will be non spontaneous if, the enthalpy is positive (endothermic reaction) and the ΔS < 0 (disorder decreases). It will not occur if we do not give energy.

ΔG < 0 → Spontaneous reaction

ΔG > 0 → Non spontaneous reaction

ΔG = 0 → System in equilibrium

8 0

3 years ago

Determine the molarity for each of the following solutions A. 1.8×10^4mg of HCL in 0.075L of solutions

Goryan [66]

The answer would have to be a

4 0

2 years ago