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3 years ago

_AlBr3+_K2SO4>_KBR+_Al2(SO4)3

Chemistry

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

Balancing chemical equation means making a number of atoms or molecules equal on both sides. In other words, this means that the number of atoms and molecules of each reacting element needs to be the same as the number of atoms and molecules of those elements in the product.

Our reaction is:

AlBr3 + K2SO4 -> KBr + Al2(SO4)3

and we need to balance it.

Since there are 3 molecules of SO4 in the product we need to put 3 before the reactant K2SO4. There are also 2 atoms of Al in the product, so we need to put 2 in front AlBr3. Now we have 6 atoms of K and Br on the left side, so we need to put 6 in front of KBr in the product.

So, our balanced equation will look like this:

2AlBr3 + 3K2SO4 -> 6KBr + Al2(SO4)3

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Copper(II) sulfide is oxidized b molecular oxygen to produce gaseous sulfur trioxide and solid copper (II) oxide. The gaseous pr

Tasya [4]

Explanation:

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The reaction is given as:

$CuS+2O_2\rightarrow CuO(s)+SO_3(g)$

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The reaction is given as:

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3 years ago

What happens when copper nitrate II is heated

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Answer:

Copper nitrate is a blue coloured crystal.which copper nitrate is heated it loses the water molecule and then when is further heated it decomposes to give redddish brown gas,nitrogen dioxide and oxygen.

Explanation:

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3 years ago

Sort each of the following elements by effective nuclear charge, Zeff, from smallest to largest:

krek1111 [17]

Answer:

Rb = +1 , Sr = +2, In= +3, Sn = +4, Sb= +5

Explanation:

Formula:

Zeff = Z - S

Z = atomic number

S = number of core shell or inner shell electrons

For Sn:

Electronic configuration:

Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²

Zeff = Z - S

Zeff = 50 - 46

Zeff = +4

For Rb:

Electronic configuration:

Rb₃₇ = [Kr] 5s¹

Zeff = Z - S

Zeff = 37 - 36

Zeff = +1

For Sb:

Electronic configuration:

Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³

Zeff = Z - S

Zeff = 51 - 46

Zeff = +5

For In:

Electronic configuration:

In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹

Zeff = Z - S

Zeff = 49 - 46

Zeff = +3

For Sr:

Electronic configuration:

Sr₃₈= [Kr] 5s²

Zeff = Z - S

Zeff = 38 - 36

Zeff = +2

8 0

3 years ago

What fraction of a Sr-90 sample remains unchanged after 87.3 years

jolli1 [7]

The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
 $t_{1/2}$ = 28.8 years

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{87.3 years}{28.8 years}$
⇒ $n=3.03$
⇒ n ≈ 3

Now we can use the first equation to calculate the remained amount of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^3$
⇒ $x= \frac{1}{8}$

8 0

3 years ago