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zubka84 [21]
3 years ago
9

Can someone help with b and c ??! Real answers :) 50 POINTS ;)

Chemistry
1 answer:
sammy [17]3 years ago
8 0

Answer:

A cell's standard state potential is the potential of the cell under standard state conditions, which is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at 25oC. Write the oxidation and reduction half-reactions for the cell.

Explanation:

sorry i could only answer B.

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Delicious77 [7]
A dust particle with mass of 0.050 and a charge
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2 years ago
Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(
Romashka [77]

<u>Answer:</u> The theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CoCl_2.6H_2O = 4.00 g

Molar mass of CoCl_2.6H_2O = 238 g/mol

Putting values in equation 1, we get:

\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol

The chemical equation for the reaction of CoCl_2.6H_2O to form  [Co(NH_3)_4(H_2O)_2]Cl_2 follows:

CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O

By Stoichiometry of the reaction:

1 mole of CoCl_2.6H_2O produces 1 mole of [Co(NH_3)_4(H_2O)_2]Cl_2

So, 0.0168 moles of CoCl_2.6H_2O will produce = \frac{1}{1}\times 0.0168=0.0168mol of [Co(NH_3)_4(H_2O)_2]Cl_2

Now, calculating the mass of [Co(NH_3)_4(H_2O)_2]Cl_2 from equation 1, we get:

Molar mass of [Co(NH_3)_4(H_2O)_2]Cl_2 = 234 g/mol

Moles of [Co(NH_3)_4(H_2O)_2]Cl_2 = 0.0168 moles

Putting values in equation 1, we get:

0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g

To calculate the percentage yield of [Co(NH_3)_4(H_2O)_2]Cl_2, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 1.20 g

Theoretical yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 3.93 g

Putting values in above equation, we get:

\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%

Hence, the theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

6 0
3 years ago
What mass of Na2SO4 is needed to make 2.5L of 2.0 M solution?
Gala2k [10]
Moles of Na2SO4 = conc. x volume(dm3)
= 2 x (2500cm3/1000)
= 5 mol


Mass of Na2SO4 = moles x molar mass
= 5 x ((23 x 2) + 32.1 + (16 x 4))
= 710.5g
5 0
3 years ago
Which sphere is dependent on all the other spheres in order to exist?
egoroff_w [7]

Answer:

biosphere

Explanation:

Living things need water (hydrosphere), chemicals from the atmosphere, and nutrients gained by eating things in the biosphere.

6 0
3 years ago
Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).
likoan [24]

Answer:

1.40*10⁻² M

Explanation:

We have the solubility formula

Solubility,

S = KH*P  

where

KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm

P = atmospheric pressure = 0.400 atm

Hence, we have

S = KH*P

= (3.50*10⁻² mol/L.atm)*(0.400 atm)

= 1.40*10⁻² mol/L

But 1 mol/L = 1 M,

Hence, the answer (1.40*10⁻² mol/L ) is equivalent to

= 1.40*10⁻² M

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3 years ago
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