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4 years ago

A nonmetallic element x forms an ionic compound with sodium that has the formula na2x. the charge on the x ion must be

1 answer:

8 0

The charge is 2- (2 minus). This is because sodium ion has a charge of 1+ and 2 of them are there in the compound. Thus the charge is 2-

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What two terms are used to describe the temperature at which solids and liquids of the same substance exist at the same time?

goldenfox [79]

Answer:

The answer is melting and freezing point.

Explanation:

5 0

3 years ago

How many nitrogen atoms are present in 0.150 mol of zn(no3)2?

gregori [183]

First calculate number of Nitrogen atoms contained by 1 mole of Zn(NO₃)₂,

As one formula unit of Zn(NO₃)₂ contains 2 Nitrogen atoms, then one mole of Zn(NO₃)₂ will contain,

= 2 × 6.022 × 10²³

= 1.20 × 10²⁴ Atoms of Nitrogen / mole

Now, calculate number of Nitrogen Atoms in 0.150 moles of  Zn(NO₃)₂.
As,
1 mole of  Zn(NO₃)₂ contained = 1.20 × 10²⁴ atoms of N
Then,
0.15 mole of  Zn(NO₃)₂ will contain = X atoms of N
Solving for X,

X = (0.15 mol × 1.20 × 10²⁴ atom) ÷ 1 mol

X = 1.80 × 10²⁴ Atoms of Nitrogen
Result:
So, 0.15 mole of  Zn(NO₃)₂ contains 1.80 × 10²⁴ atoms of Nitrogen.

8 0

3 years ago

Which sentence describes a star clusters ? Check all that apply

Sholpan [36]

B. stars form large groupings

3 0

3 years ago

Read 2 more answers

Calculate the molarity of a solution made by adding 45.4 g of nano3 to a flask and dissolving it with water to create a total vo

aleksandr82 [10.1K]

Solutions are made up of two non reacting species called solute and solvent. The amount of solute in solvent is known as concentration of that solute. Concentration is often measured in Molarity. Molarity is the amount of solute dissolved in 1 dm3 of solution. Answer to your question is as follow;

3 0

3 years ago

Read 2 more answers

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points

labwork [276]

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

$Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

$NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

$H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

$NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$

The net ionic equation is as follows.

$H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

7 0

3 years ago