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Lilit [14]
3 years ago
14

A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask

to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
Chemistry
1 answer:
leonid [27]3 years ago
8 0

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer:  The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of Ba(CH_3COO)_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.125\times 1000}{350ml}

Molarity=0.357M

Therefore, the concentration of solution is 0.375 mol/L

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Ag(s)+CN−(aq)+O2(g)→Ag(CN)−2(aq) Express your answer as a chemical equation. Identify all of the phases in your answer. Express
Lerok [7]

Answer : The balanced redox reaction will be,

4Ag(s)+8CN^-(aq)+O_2(g)+4H^+(aq)\rightarrow 4Ag(CN)_2^-(aq)+2H_2O(l)

Explanation :

The given chemical equation is,

Ag(s)+CN^-(aq)+O_2(g)\rightarrow Ag(CN)_2^-(aq)

In the half reaction method, the number of atoms in each half reaction and number of electrons must be balanced.

The half reactions in the acidic medium are :

Reduction : O_2(g)+4H^+(aq)+4e^-\rightarrow 2H_2O(l)   ......(1)

Oxidation : Ag(s)+2CN^-(aq)\rightarrow Ag(CN)_2^-(aq)+1e^-  .......(2)

Now multiply the equation (2) by 4 and then added both equation, we get the balanced redox reaction.

Thus, the balanced redox reaction will be,

4Ag(s)+8CN^-(aq)+O_2(g)+4H^+(aq)\rightarrow 4Ag(CN)_2^-(aq)+2H_2O(l)

7 0
4 years ago
At what celsius temperature does 0.750mol of an ideal gas occupy a volume of 35.9L at 114kPa
pashok25 [27]

Answer:

378.25°C

Explanation:

Given data:

Number of moles of gas = 0.750 mol

Volume of gas = 35.9 L

Pressure of gas = 114 KPa (114/101 = 1.125 atm)

Temperature of gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

now we will put the values.

T = PV/nR

T = 1.125 atm × 35.9 L /0.750 mol  × 0.0821 atm.L/ mol.K

T = 40.3875/0.062/K

T = 651.4 K

Kelvin to °C:

651.4 K - 273.15 = 378.25°C

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3 years ago
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