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3 years ago

How test berween nitric and sulferic acid?

1 answer:

7 0

Ba(OH)₂(aq) + 2HNO₃(aq) = Ba(NO₃)₂(aq) + 2H₂O(l)
At the reaction of barium hydroxide with nitric acid there are no visual changes.

Ba(OH)₂(aq) + H₂SO₄(aq) = BaSO₄(s) + 2H₂O
At the reaction of barium hydroxide with sulfuric acid the precipitat of white color is formed.

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Which of the following elements would be likely to form an anion?

Vsevolod [243]

B because chlorine is a halogen. This means that it already 7 valence electrons and only needs 1 more to have a complete outer shell.

4 0

3 years ago

A wide range of molecular compounds dissolve in water. this allows water to act as a medium for moving molecules into and out of

Sergio [31]

Lipids cannot dissolve in water.

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3 years ago

Read 2 more answers

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

5 0

3 years ago

How does changing the ratio of nitrogen atoms and oxygen atoms change the compounds

Reptile [31]

Answer:

Explanation:

Compounds consist of atoms of different elements. In compound the atoms are present in fixed proportion. By changing the proportion compound will changed. For example,

Nitrogen and oxygen reacted to form the compound. By changing the proportion both will form different compound.

NO₂

Nitrogen dioxide consist of one atom of nitrogen and two atoms of oxygen. Their ration is 1 :2

In nitric oxide one atom of oxygen and one atom of nitrogen are present.

Their ratio is 1 : 1.

N₂O

Nitrous oxide consist of two nitrogen atoms and one oxygen. Their ratio is

2 : 1

N₂O₄

Dinitrogen tetroxide consist of two nitrogen atoms and four oxygen atoms. Their ratio will written as 2 : 4

N₂O₅

Dinitrogen pentoxide consist of two nitrogen atoms and five oxygen atoms. Their ratio will written as 2 : 5

It can be seen that by changing the ratio of atoms compounds are changed.

8 0

3 years ago

A sample of an ideal gas at 1.00 bar and a volume of 1.59 l was placed in a weighted balloon and dropped into the ocean. as the

dimulka [17.4K]

Le us discuss how we should approach this problem. We know that there are two states. The first state is when P₁ = 1 bar and V₁ = 1.59 L. The second state is when P₂ = 80 bar. We are asked to find V₂. At constant temperature, we use Boyle's Law.

P₁V₁ = P₂V₂
V₂ = P₁V₁/P₂ = (1 bar)(1.59 L)/(80 bar)
<em>V₂ = 0.02 L</em>

8 0

3 years ago