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Studentka2010 [4]
2 years ago
8

Of the following substances, an aqueous solution of ________ will form basic solutions. nh4cl cu(no3)2 k2co3 naf a) k2co3, nh4cl

b) naf only c) nh4cl only d) naf, k2co3 e) nh4cl, cu(no3)2
Chemistry
1 answer:
Andrew [12]2 years ago
6 0

The correct option is (a) NaHS, KHCO3 and NaF

NaHS, KHCO3 and NaF will form an aqueous solution of form basic.

<h3>What is aqueous solution?</h3>

A solution in which water serves as the solvent is called an aqueous solution. The most common way to display it in chemical equations is by adding to the appropriate chemical formula. For instance, the symbol for a solution of table salt, or sodium chloride, in water is Na+ + Cl.

To learn more about aqueous solution visit:

brainly.com/question/14097392

#SPJ4

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0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
3 years ago
Add oxidation numbers to the following reaction: 2 H3PO4 (aq) + 2 Cr(s) → 2 CrPO4 (aq) + 3 H2(g). Identify the atom that is oxid
rewona [7]

The atom that is oxidized : Cr

The oxidizing agent : H₃PO₄

<h3>Further explanation</h3>

Reaction

2 H₃PO₄ (aq) + 2Cr(s) → 2 CrPO₄ (aq) + 3H₂(g)

Atoms undergoing a reduction reaction (decrease in oxidation number) and an oxidation reaction (increase in oxidation number)

  • Reduction (+1 to 0)

H⁺(in H₃PO₄) =+1

H₂=0

  • Oxidation (0 to +3)

Cr = 0

Cr³⁺(in CrPO₄ )

the oxidizing agent.⇒which undergoes a reduction reaction and oxidizes another compound/element : H₃PO₄

3 0
3 years ago
Calculate the ph of the resulting solution if 27.0 ml of 0.270 m hcl(aq is added to 37.0 ml of 0.270 m naoh(aq
bearhunter [10]
The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first 
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63
3 0
3 years ago
Read 2 more answers
How to do part (b)??
laila [671]

it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.

3 0
3 years ago
Write a balanced half-reaction for the reduction of gaseous oxygen to aqueous hydrogen peroxide in basic aqueous solution. Be su
Vesna [10]

Answer:

Balanced half-reaction: O_{2}(g)+2H_{2}O(l)+2e^{-}\rightarrow H_{2}O_{2}(aq.)+2OH^{-}(aq.)

Explanation:

Balanced half-reaction for the reduction of gaseous oxygen to aqueous hydrogen peroxide in basic solution should be written in the following step-by-step manner:

Half-reaction: O_{2}(g)\rightarrow H_{2}O_{2}(aq.)

Balance H and O in basic medium: O_{2}(g)+2H_{2}O(l)\rightarrow H_{2}O_{2}(aq.)+2OH^{-}(aq.)

Balance charge: O_{2}(g)+2H_{2}O(l)+2e^{-}\rightarrow H_{2}O_{2}(aq.)+2OH^{-}(aq.)

Balanced half-reaction: O_{2}(g)+2H_{2}O(l)+2e^{-}\rightarrow H_{2}O_{2}(aq.)+2OH^{-}(aq.)

7 0
4 years ago
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