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Misha Larkins [42]
3 years ago
7

When atoms of different elements combine, the new molecules are called: A. matter B. particles C. compounds D. mass

Chemistry
1 answer:
WITCHER [35]3 years ago
7 0
C is correct answer.


Compounds is a atoms it has a different elements combine the new molecules.


Hope it helped you.

-Charlie
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What is the molarity of a sodium chloride solution made by dissolving 4.512 moles to make 2.0 L?
Svet_ta [14]

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

mL

. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

1 L

=

10

3

mL

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

1

, which means that you're going to have to convert the mass of the sample from milligrams to grams

1 g

=

10

3

mg

Sodium chloride,

NaCl

, has a molar mass of

58.44 g mol

−

1

, which means that your sample will contain

unit conversion



280.0

mg

⋅

1

g

10

3

mg

⋅

molar mass



1 mole NaCl

58.44

g

=

0.004791 moles NaCl

This means that the molarity of the solution will be

c

=

n

solute

V

solution

c

=

0.004791 moles

2.00

⋅

10

−

3

L

=

2.40 M

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.

6 0
3 years ago
What does chemistry have to do with drug testing
diamong [38]

Answer:

Forensic drug chemists analyze samples of unknown materials including powders, liquids and stains to determine the chemical identity or characteristics of the compounds that make up the sample. samples submitted as evidence in a drug-related case can contain one compound or a mixture of many compounds.

5 0
3 years ago
Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
GalinKa [24]
<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0
3 years ago
A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±
Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0
3 years ago
Using the reaction C12H22O11 + 12 O2 -&gt;12CO2 + 11H2O (e). How many grams of carbon dioxide can be produced from 456.7g sucros
klio [65]

Answer:

704.6 g CO2

Explanation:

MM sucrose = 342.3 g/mol

MM CO2 = 44.01 g/mol

g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2

7 0
2 years ago
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