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2 years ago

An electron ______ is an area around the nucleus of an atom where an electron is likely to be found.

Chemistry

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

orbital

hope this helps! <3

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Explain what it means by a balanced chemical equation

larisa86 [58]

It's balanced if the products and the reactants have exactly the same number of atoms

8 0

2 years ago

Calculate the number of atoms in 0.316 mol of phosphorus

Nikitich [7]

Answer:

$\large \boxed {7.61 \times 10^{23}\text{ atoms}}$

Explanation:

1. Molecules of P₄

The molecular formula of phosphorus is P₄.

$\text{Molecules of P}_{4} = \text{0.316 mol P}_{4} \, \times \dfrac{ 6.022 \times 10^{23}\text{ molecules P}_{4}}{\text{1 mol P}_{4}} = 1.90 \times 10^{23}\text{ molecules P}_{4}$

2. Atoms of P

One molecule of phosphorus contains four P atoms.

$\text{Atoms of P} = 1.90 \times 10^{23}\text{ molecules P}_{4} \times \dfrac{\text{4 atoms P}}{\text{1 molecule P}_{4}}\\\\= \large \boxed{\mathbf{7.61 \times 10^{23}}\textbf{ atoms P}}$

5 0

2 years ago

Please use the values in the resources listed below instead of the textbook values. Under certain conditions the decomposition o

Alex787 [66]

Answer:

The rate equation for this reaction:

$R=k[NH_3]^0$

Explanation:

Decomposition of ammonia:

$2NH_3\rightarrow N_2+3H_2$

Rate law of the can be written as;

$R=k[NH_3]^x$

1) Rate of the reaction , when $[NH_3]=2.0\times 10^{-3} M$

$1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x$ ..[1]

2) Rate of the reaction , when $[NH_3]=4.0\times 10^{-3} M$

$1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x$ ..[2]

[1] ÷ [2]

$\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}$

On solving for x , we get ;

x = 0

The rate equation for this reaction:

$R=k[NH_3]^0$

4 0

3 years ago

If a reaction mixture initially contains 0.195 m so2cl2, what is the equilibrium concentration of cl2 at 227 ???c?

Aloiza [94]

Missing in your question the K constant value =2.99X10^-7
The reaction equation:
SOCl2 ⇄ SO2 +Cl2
initial 0.195 0 0
-X +X +X
final (0.195-X) X X

K= [SO2][Cl2]/SOCl2
by substitution
2.99X10^-7 = (X)(X)/(0.195-X)
2.99X10^-7 = X^2 / (0.195-X) by solving this equation
∴X = 0.00024 = 2.4 X10^-4
∴[Cl2] = 2.4X10^-4 m

6 0

3 years ago

How many grams is 6.31 x 1024 molecules of BF3 in scientific notation?

zhannawk [14.2K]

Answer:

6,461.44 sana po makatulong

5 0

2 years ago