All categories

4 years ago

Is a diamond made in a laboratory a mineral

Chemistry

2 answers:

mezya [45]4 years ago

6 0

No because mineral must be naturally occurring,if it is made in the laboratory is considered synthetic

Nikolay [14]4 years ago

3 0

I think it is but it is artificial but still real and is still a mineral
hope this helps ∞

You might be interested in

How many moles of Hcl can be produced from 1.21 moles of zn?

Gre4nikov [31]

Answer:

2 moles

Explanation:

5 0

3 years ago

g A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the

ElenaW [278]

Answer:

Answer: 882mmHg

Explanation:

In a closed-end manometer, the gas added is under a pressure that is measured against atmospheric pressure. The difference in pressure between both gases is equal to the difference in mercury levels. Thus, the pressure of the argon can be:

783 mmHg - 96.0mmHg = 687mmHg

783 mmHg - 96.0mmHg = 879mmHg

Based on the options, the possible answer (The nearest to the second value) is:

<h3>Answer: 882mmHg</h3>

7 0

3 years ago

What is the maximum mass of s8 that can be produced by combining 89.0 g of each reactant?

lidiya [134]

Answer : We can produce 125.7 g of $S_{8}$ .

Explanation : The reaction will be

$8SO_{2} + 16H_{2}S -----> 3S_{8} + 16H_{2}O$

The molecular mass of $SO_{2}$ is 64.1 g/mol

and molecular mass of $H_{2}S$ is 34.1 g/mol

For every mole of $SO_{2}$ we would need twice of $H_{2}S$ moles, so for every 3 moles of $S_{8}$ we need 16 moles of $H_{2}S$

Now, we can calculate number of moles $S_{8}$

2.61 X (3/16) = 0.49 moles

Here, the molecular mass of $S_{8}$ is 256.8 g

multiplying it with the number of 0.49 moles we get, 256.8 X 0.49 = 125.7 g of $S_{8}$ .

Hence, 125.7 g of $S_{8}$ will be produced.

8 0

3 years ago

What is the molarity of a solution prepared by dissolving 36. 0 g of naoh in enough water to make 1. 50 l of solution?

Vlada [557]

0.6 mol / L is the molarity of a solution prepared by dissolving 36. 0 g of NaOH in enough water to make 1. 50 l of solution.

The amount of a substance in a specific volume of solution is known as its molarity (M). The number of moles of a solute per liter of a solution is known as molarity. The molar concentration of a solution is another term for molarity.

The ratio employed to indicate the solution's concentration is called its molarity. Understanding a solution's molarity is important since it allows you to determine the actual concentration as well as whether the solution is diluted or concentrated.

Amount of NaOH = 36. 0 g

Amount of water = 1. 50 L

1 mol of NaOH = 40 g,

Moles of NaOH = 36. 0 / 40 g = 0.9 mol NaOH

Molarity of a solution = moles of solute / Liters of solution

Molarity of a solution = 0.9 / 1.50

Molarity of a solution = 0.6 mol / L

To know more about Molarity refer to: brainly.com/question/8732513

#SPJ4

4 0

1 year ago

The combustion of a 0.4255 g sample of a compound containing carbon, hydrogen, oxygen and bromine produces 0.4961 g CO2 and 0.17

IgorC [24]

Answer: The empirical formula is C11HO3Br

Explanation: 1ST SAMPLE;

First, we need to get the number of moles;

Molar mass of C=12, O=16, H=1, Ag =107.9, Br=79.9

0.4961g CO2 x (1 mol CO2) / (44.0g CO2) = 0.0113 mol of Co2

Since one mole of CO2 is made up of 1 mole of C and 2 moles of O, and we have 0.0113 moles of CO2 in our sample, then we know we have 0.0113 moles of C in the sample. Now, we need to know the mass of C we have in the sample:

(0.0113 mol C) (12.011g C/1 mol C) = 0.136g C

Now we follow the same pattern above and do it for the hydrogen:

0.1777g H2O x (1 mol H2O) / (18g H2O) = 0.01 mol

Now, for the mass of H:

(0.01 mol H) (1g H/1 mol H) = 0.01g H

But this is for 1 mole of H. Whereas, H has 2 moles from the question, therefore it's equal to 0.01 x 2 = 0.02g H

Since we combusted 0.4255g of the sample, the missing mass will be from the bromine and oxygen since we have gotten masses of Carbon and Hydrogen.

Therefore missing mass = 0.4255 - (0.136+0.02) = 0.2695 of bromine and oxygen

2ND SAMPLE:

First, we need to get the number of moles;

0.1894 g AgBr x (1 mol AgBr) / (107.9 + 79.9g AgBr) = 0.001 mol of AgBr

Since AgBr is made up of 1 mole of Ag and 1 mole of Br each,

We can say that there's 0.001 mole of Br in this second sample.

Now looking at the first and second samples, lets set up a proportion to know the number of moles of Br in the first sample.

If 0.1523g of the second sample produced 0.001 mol of Br, therefore 0.4255g in the first sample will produce: (0.4255g x 0.001)/0.1523 = 0.0028mol of Br

Therefore the mass of Br in the first sample is: (0.0028 mol C) (79.9g Br/1 mol C) = 0.22372g Br

From the first sample, we saw that the sum of Br and Oxygen equals 0.2695

Therefore,the mass of oxygen is: 0.2695 - 0.22372 = 0.04578g of oxygen

Therefore to find number of moles of oxygen;

(0.04578g O x (1 mol O) / (16.0g O) = 0.0029 mol of oxygen

Overall, we have; C=0.0113 moles ;H=0.001 moles; Br= 0.0028moles and O= 0.0029

The smallest is 0.001. So to simplify this for the empirical formula, we divide each by 0.001 to get approximately C= 11, H=1, Br=3, O=3

Therefore the empirical formula is C11HO3Br

5 0

3 years ago