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sladkih [1.3K]
3 years ago
14

Three teaspoons of sucrose(17.1 g) is dissolved in hot water to make a total volume of 250 ml solution. Calculate the molarity

Chemistry
1 answer:
dusya [7]3 years ago
5 0
Molar mass sucrose :

C12H22O11 =  342.29 g/mol

number of moles <span>of the solute :
</span>
n = mass solute / molar mass

n = 17.1 / 342.29

n = 0.05 moles

Volume in liters :

250 mL / 1000 =  0.25 L


therefore :

Molarity  =  moles of solute / volume

Molarity =  0.05 / 0.25  => 0.2 mol/L


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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the
lutik1710 [3]

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

3 0
3 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
Kryger [21]

Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

8 0
3 years ago
Can someone help me, please?
DENIUS [597]
The answer to this question is: B) 3+
7 0
3 years ago
Read 2 more answers
What mass of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
iren [92.7K]

91.4 grams

91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

C = mol/volume

2.45M=mol/0.5L

2.45M⋅0.5L = mol

mol = 1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225mol * \frac{39.1+35.5}{mol}

mol=1.225

=1.225 mol . \frac{74.6g}{mol}

=1.225 . 74.6

=91.4g

therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

What is 1 molar solution?

In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.

58.44 g make up a 1M solution of NaCl.

To learn more about molar solution visit:

brainly.com/question/10053901

#SPJ4

3 0
1 year ago
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