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Norma-Jean [14]
3 years ago
13

Why is the concentration of 1.34gvCuCl2 in a volumetric flask with approx. 100mL of water has a concentration of 0.050M?

Chemistry
1 answer:
jarptica [38.1K]3 years ago
8 0
<h3>Answer:</h3>

Concentration will be 0.1 M and not 0.050M

<h3>Explanation:</h3>

This explains why the molarity is not 0.050 M but 0.1 M

Molarity is the concentration of a solution in moles per liter.

Therefore,

Molarity = moles ÷ Volume

In this case,

We have 1.34 g of CuCl₂ and a volume of 100 mL

To get the molarity we need to find the number of moles in 1.34 CuCl₂

Number of moles = Mass÷ molar mass

Molar mass of CuCl₂ = 134.45 g/mol

Therefore,

Moles of CuCl₂ = 1.34 g ÷ 134.45 g/mol

                         = 0.009965 moles

                         = 0.01 moles

Then we can calculate the molarity;

Molarity = Number of moles ÷ Volume

Volume of the solution is 100 mL or 0.1 L

Molarity = 0.01 moles ÷ 0.1 L

            = 0.1 M

Therefore, the concentration of 1.34 g CuCl₂ in a volume of 100 mL is 0.1 M and not 0.050M.

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How many mL of 4.0% Mg(NO3)2 solution would contain 1.2 g of magnesium nitrate?
yulyashka [42]
4.0 % in mass per volume :

4.0 g ---------- 100 mL
1.2 g ----------  ( volume )

Volume =  1.2 x 100 / 4.0

Volume = 120 / 4.0

= 30 mL

hope this helps!
5 0
3 years ago
The empirical formula for trichloroisocyanuric acid
Naddika [18.5K]

Answer:

OCNCl

Explanation:

The empirical formula for trichloroisocyanuric acid, the active ingredient in many household bleaches, is OCNCl. The molar mass of this compound is 232.41g/mol.

7 0
3 years ago
Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M)
katrin2010 [14]

<u>Answer:</u> The electrode potential of the cell is 0.093 V

<u>Explanation:</u>

The given chemical cell follows:

Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)

<u>Oxidation half reaction:</u> H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V

<u>Reduction half reaction:</u> Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V

<u>Net cell reaction:</u>

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.13-(0.0)=0.13V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

[H^{+}]=1.00M

[Sn^{2+}]=0.350M

[Sn^{4+}]=0.020M

Putting values in above equation, we get:

E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})

E_{cell}=0.093V

Hence, the electrode potential of the cell is 0.093 V

3 0
4 years ago
Which is a waste product of aerobic respiration? A. water B. energy C. oxygen D. glucose
Svetradugi [14.3K]

Answer: A

Explanation:

7 0
4 years ago
My car has an internal volume of 3,600 L. If the sun heats my car from a temperature of 18.0 degrees Celsius to a temperature of
Papessa [141]

Answer:

110.7 kpa or 1.097 atm your choice of pressure units

Explanation:

We can assume the Volume is constant since we are not told other wise and it is indeed a car. We can Use P1/T1= P2/T2

So (T2xP1 )/ T1 = P2

(318Kx101.3kpa)/291k = 110.7 kpa or 1.097 atm your choice of pressure units

if your wondering why i used Kelvin gas laws are always in units of Kelvin

5 0
3 years ago
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