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2 years ago

Determine the mass of 4.1 x 1023 atoms Cl.Answer in units of g

Chemistry

1 answer:

icang [17]2 years ago

8 0

Answer:24.14g Cl

Explanation:

Convert to moles

4.1 x 10^23 atoms Cl x (1 mol/6.022 x 10^23 stoms) = 0.68 moles

Convert from moles to grams using the molar mass by periodic table

0.68moles(35.45g/1 mol) = 24.14g Cl

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in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola

Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for Volume

V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0

2 years ago

PLEASE HELP!!!!!!!

dsp73

Answer: 1. $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

2. 3 moles of $CuCl_2$ : 2 moles of $Al$

3. 0.33 moles of $CuCl_2$ : 0.92 moles of $Al$

4. $CuCl_2$ is the limiting reagent and $Al$ is the excess reagent.

5. Theoretical yield of $AlCl_3$ is 29.3 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles$

$\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles$

The balanced chemical equation is:

$2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

According to stoichiometry :

3 moles of $CuCl_2$ require = 2 moles of $Al$

Thus 0.33 moles of $CuCl_2$ will require= $\frac{2}{3}\times 0.33=0.22moles$ of $Al$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $CuCl_2$ give = 2 moles of $AlCl_3$

Thus 0.33 moles of $CuCl_2$ give = $\frac{2}{3}\times 0.33=0.22moles$ of $AlCl_3$

Theoretical yield of $AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3$

Thus 29.3 g of aluminium chloride is formed.

5 0

2 years ago

Practice It!

Paladinen [302]

Mitosis hope that helps i tried:)

4 0

2 years ago

Read 2 more answers

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.