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andreyandreev [35.5K]
3 years ago
5

"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi

s 25.0 mL sample with water until the new volume equals 600.0 mL, what is the concentration of nitrate ions in the 600.0 mL solution
Chemistry
1 answer:
ycow [4]3 years ago
7 0

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

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A 10.00 g sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 g BaSO4 (M- 233.4). W
aliina [53]

Answer:

The salt is barium chloride.

Explanation:

BaX_2++Na_2SO_4\rightarrow BaSO_4+2NaX

Moles of barium sulfate =\frac{11.21 g}{233.38 g/mol}=0.0480 mol

According to reaction, 1 mol of barium sulfate is produced from 1 mol of BaX_2.

Then 0.0480 moles will be produced from:

\frac{1}{1}\times 0.0480 mol=0.0480 mol of BaX_2.

Mass of BaX_2 used = 10.00 g

Moles of BaX_2 =\frac{10.00 g}{\text{Molar mass}}[/tex]

0.0480 mol=\frac{10.00}{\text{Molar mass}}

Molar mass of BaX_2 = 208.33 g/mol

The nearest answer to our answer is BaCl_2=208.2 g/mol.

The correct answer barium chloride with molar mass of 208.2 g/mol.

8 0
3 years ago
b → p experiment number [a] (m) [b] (m) initial rate (m/s) 1 0.273 0.763 2.83 2 0.273 1.526 2.83 3 0.819 0.763 25.47 22) the rat
Mazyrski [523]

The rate law for this reaction is [A]².

Balanced chemical reaction used in this experiment: A + B → P

The reaction rate is the speed at which reactants are converted into products.

Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.

Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².

More info about rate law: brainly.com/question/16981791

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In the alveoli and lung capillaries, co2 and o2 are exchanged by means of _____.
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Through hemoglobin transport in blood, material exchange is carried out on the alveolar membrane.
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2 years ago
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
3 years ago
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