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galben [10]
3 years ago
13

The atomic mass number of aluminum is 26.98. If you had 1.34 x 1025 atoms of aluminum, how much would they all weigh?

Chemistry
1 answer:
a_sh-v [17]3 years ago
3 0

Answer:

601 gram

Explanation:

1st we should know how may moles are there in 1.34 X 10 25 atoms

 # moles of Al = # of atoms / # of Avogadro's num

                      = 1.34 X 10 25/ 6.12X10 23 =

                       = 22.26 mol of AL

 so mass of AL = moles x m.wt

                          =22.26 x 26.98 = 600.55 = 601 gram

DONE

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A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
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Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

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Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

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substituting for k,t and [a] we get:

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Naily [24]

Answer:

A: 2.525 x 10-4 mol

B: 2.583 x 10-4 mol

Explanation:

Part A:

Data Given:

. Temperature of water (H2O) = 21.3°C

Convert Temperature to Kelvin

T = °C + 273

T = 21.3 + 273 = 294.3 K

volume of (H2O) gaseous state = 5.1 mL

Convert mL to liter

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Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n=PV/RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/ mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0051 L / 0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0061 atm.L / 24.162 L.atm.mol-1

n = 2.525 x 10-4 mol

no. of moles of gas (H2O) = 2.525 x 10-4 mol

Part B:

Data Given:

Temperature of water (H2) = 21.3°C

Convert Temperature to Kelvin

T = "C + 273

T= 21.3 + 273 = 294.3 K

volume of (H2) gas = 5.2 mL

Convert mL to liter

1000 mL = 1 L

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Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n= PV / RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0052 L/0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0062 atm.L/ 24.162 L.atm.mol-1

n = 2.583 x 10-4 mol

I

no. of moles of gas (H2) = 2.583 x 10-4 mol

8 0
3 years ago
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