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Temka [501]
3 years ago
15

Write the electron configuration​

Chemistry
1 answer:
ycow [4]3 years ago
8 0

Answer:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
2 years ago
If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is
Studentka2010 [4]

Answer:

<em>(H30+)= 1x10^-6 M</em>

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

<em>(H30+)= 1x10^-6 M</em>

6 0
3 years ago
Is gravity a matter??
algol13

Answer:

No, gravity isn't matter

Explanation:

Gravity is a <u>force</u> that attracts matter towards the center of a physical body with mass.

3 0
2 years ago
Calculate the number of moles of H2 produced in the reaction of Mg(s) with HCl(aq). Mg(s) is the
Taya2010 [7]

Explanation:

Moles of metal,

=

4.86

⋅

g

24.305

⋅

g

⋅

m

o

l

−

1

=

0.200

m

o

l

.

Moles of

H

C

l

=

100

⋅

c

m

−

3

×

2.00

⋅

m

o

l

⋅

d

m

−

3

=

0.200

m

o

l

Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.

So if

0.200

m

o

l

acid react, then (by the stoichiometry), 1/2 this quantity, i.e.

0.100

m

o

l

of dihydrogen will evolve.

So,

0.100

m

o

l

dihydrogen are evolved; this has a mass of

0.100

⋅

m

o

l

×

2.00

⋅

g

⋅

m

o

l

−

1

=

?

?

g

.

If 1 mol dihydrogen gas occupies

24.5

d

m

3

at room temperature and pressure, what will be the VOLUME of gas evolved?

5 0
2 years ago
Can someone help me with this molar mass problem?[It’s the last one]
quester [9]

Answer:

54.18 \times 10^{23} \ moles in 3 mole of  Al_2(SO_4)_3

Explanation:

It is clear that in the given 1\ mole of Al_2(SO_4)_3 have 3\ ions of SO_4^2^-

Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Since 1 ion of anything is equivalent to 6.02\times10^{23} \ moles

Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Which is equivalent to 9 \times6.02\times10^{23}=54.18\times10^{23} \ moles

Thus 3 moles of  Al_2(SO_4)_3 gives 54.18\times10^{23} \ moles of  SO_4^2^-.

5 0
3 years ago
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