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Jobisdone [24]
3 years ago
5

How many moles of AIC3 are there in 1,119.972 g ?

Chemistry
1 answer:
Alchen [17]3 years ago
6 0

Answer:

it is 8.40189 moles of AlCl3

mole = mass/molar mass

mole= 1119.972/133.34

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This is the answer because it is the only way that it is okay to split in half.

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A student places a beaker on a hot plate. To the beaker, he adds a clear liquid and a white powder. The mixture instantly begins
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100 ml is drawn from 0.1 M solution of KCl and added to 900 ml of water. What is the
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Answer:

The new concentration will be 0.01 M.

Explanation:

To determine the new concentration we use the following formula.

concentration (1) × volume (1) = concentration (2) × volume (2)

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volume (1) = 100 mL

concentration (2) = unknown

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3 0
3 years ago
What is an atomic number?
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Consider the equation:
Dmitry [639]

Answer:1. Rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}

2. The rate constant (k) for the reaction is 3.50M^\frac{-1}{2}s^{-1}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

rate=k[CHCl_3]^x[Cl_2]^y

k= rate constant

x = order with respect to CHCl_3

y = order with respect to Cl_2

n = x+y= Total order

1. a) From trial 1: 0.0035=k[0.010]^x[0.010]^y  (1)

From trial 2: 0.0069=k[0.020]^x[0.010]^y   (2)

Dividing 2 by 1 :\frac{0.0069}{0.035}=\frac{k[0.020]^x[0.010]^y}{k[0.010]^x[0.010]^y}

2=2^x,2^1=2^x therefore x=1.

b)  From trial 2: 0.0069=k[0.020]^x[0.010]^y   (3)

From trial 3: 0.0098=k[0.020]^x[0.020]^y   (4)

Dividing 4 by 3:\frac{0.0098}{0.0069}=\frac{k[0.020]^x[0.020]^y}{k[0.020]^x[0.010]^y}

1.4=2^y,2^{\frac{1}{2}}=2^y therefore y=\frac{1}{2}

rate=k[CHCl_3]^1[Cl_2]^\frac{1}{2}

2. to find rate constant using trial 1:

0.0035=k[0.010]^1[0.010]^\frac{1}{2}  

k=3.50M^\frac{-1}{2}s^{-1}

5 0
4 years ago
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