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alexdok [17]
3 years ago
14

How many moles in 53g of palladium 216g of silver 46g of tungsten

Chemistry
2 answers:
Helga [31]3 years ago
7 0
To answer this, all you have to do is take the mass of the sample, in grams, and divide it by the atomic/molecular mass of the substance, in amu (atomic mass units)
So, here are my answers:
(i) 53g of Palladium = 53/106 = 0.5 moles
(ii) 216g of Silver = 216/108 = 2 moles
(iii) 46g of Tungsten = 46/184 = 0.25 moles

Hope I helped!! xx
snow_tiger [21]3 years ago
5 0
Moles = Mass / Mr 
You basically need to open the periodic table and divide the mass to find out the moles. 

Palladium = 53/ 106
                 = 0.5 moles

Silver = 216/108
          = 2 moles

Tungsten = 46/ 184
                = 0.25 moles
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By there pH . a pH below 7 is acidic . Above 7 is basic. If it’s right at 7 it’s neutral.
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The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below.
AnnZ [28]

<span>I believe the correct 2nd reaction is:</span>

cof2(g)⇌1/2 co2(g)+1/2 cf4(g)

where we can see that it is exactly one-half of the original

Therefore the new Kp is:

new Kp = (old Kp)^(1/2)

new Kp = (2.2 x 10^6)^(1/2)

<span>new Kp = 1,483.24 </span>

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3 years ago
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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


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attashe74 [19]

Answer: sorry for the late answer, I just took the test today.

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