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patriot [66]
2 years ago
6

How many milliliters of 0. 0140 m na2s2o3 solution are needed to dissolve 0. 590 g of agbr? agbr(s) 2na2s2o3(aq)→na3ag(s2o3)2(aq

) nabr(aq)
Chemistry
1 answer:
Marina86 [1]2 years ago
3 0

The molarity of the solution can be used to give the moles and the volume of the solution. The 0.0140 M sodium thiosulfate needs 225 ml of volume.

<h3>What is molarity?</h3>

Molarity is the property of the solution that is given by the ratio of the moles of the solute to the volume of the solution in liters.

Given,

Molarity of Sodium thiosulfate = 0.0140 M

Mass of solute (AgBr) = 0.590 gm

Molar mass of AgBr = 187.77 g/mol

First moles from the mass can be calculated as,

Moles = mass ÷ molar mass

n = 0.590 g ÷ 187.77 g/mol

= 0.00314 moles

Now, the volume of the solution from molarity is calculated as,

Volume = moles ÷ molarity

= 0.00314 moles ÷ 0.0140

= 0.225 L

Therefore, 225 mL is the volume required.

Learn more about molarity here:

brainly.com/question/13889022

#SPJ4

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3.25 x 10+8 nm2 divide by 6.5 x 10+6 nm =
Vesnalui [34]

Answer:  50 nm

Explanation:  Two steps:

1.  Divide 3.25/6.5 = 0.5

2.  Divide 10^8/10^6 = 10^2

nm^2/nm = nm

Combine:  0.5x10^2 nm

or 50 nm

8 0
2 years ago
A solution contains one or more of the following ions: Ag+, Ca2+, and Co2+. Lithium bromide is added to the solution and no prec
olchik [2.2K]
<h3>Answer:</h3>

#1. Ca²⁺

# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

<h3>Explanation:</h3>

The question above concerns solubility of salts or ions in water.

The solution given contains Ag+, Ca2+, and Co2+ ions.

  • In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
  • In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
  • The net ionic equation for the reaction is;

Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

  • From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
  • In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
  • The net ionic equation for the reaction is;

3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

  • According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
6 0
3 years ago
A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
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