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3 years ago

Which metal is more active than tin, but less active than zinc

Chemistry

1 answer:

VikaD [51]3 years ago

8 0

Answer:

Iron (Fe)

Explanation:

The reactivity series shows a series of metals reactivity from lowest to highest. Based on the arrangement of metals in the reactivity series, iron is placed above tin and below zinc, this means that iron is more reactive than tin but less reactive than zinc. This means that iron would displace tin ions in a solution whereas zinc would displace iron ions in a solution.

The reactivity of metals is due to the stability of metals in their electron configurations as atoms and as ions.

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g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

RUDIKE [14]

Answer:

For a: Lead iodide is a yellow precipitate.

For b: Barium sulfate is a white precipitate.

For c: Ferric hydroxide is a brown precipitate.

For d: Copper (II) hydroxide is a blue precipitate.

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

For (a):

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

For b:

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

For c:

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

For d:

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

6 0

3 years ago

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LenKa [72]

Answer:

hahahh noob

Explanation:

3 0

3 years ago

1. 7.85 g of sodium metal is added to 200 mL of 0.0450 M HNO3

kumpel [21]

Answer:

a)0.765 g

b)7.613 g

c)0.20 L

Explanation:

Consider the reaction involved;

Na(s) + HNO3(aq) ----> NaNO3(s) + H2(g)

Note that, if a hot, saturated aqueous solution of sodium nitrate was allowed to cool, solid sodium nitrate would crystallise out of the solution and this would also be classed as a precipitate. This is the case here.

Number of moles of sodium reacted= mass of sodium reacted/ molar mass of sodium

Number of moles of sodium= 7.85g/23gmol-1

Number of moles of sodium= 0.34 moles of sodium

Number of moles of acid reacted= concentration of acid × volume of acid

Number of moles of acid= 0.0450 × 200/1000

Number of moles of acid= 9×10^-3 moles

Therefore, HNO3 is the limiting reactant.

1 mole of HNO3 yield 1 mole of NaNO3

9×10^-3 moles of HNO3 yield 9×10^-3 moles of NaNO3

Hence mass of NaNO3= number of moles × molar mass

Mass of NaNO3= 9.0×10^-3 moles × 84.9947 g/mol

Mass of NaNO3= 0.765 g of NaNO3

Since

1 mole of sodium metal reacts with 1 mole of HNO3

9×10^-3 moles of sodium reacts with 9×10^-3 moles of HNO3

Therefore amount of unreacted sodium metal = 0.34 moles - 9×10^-3 moles = 0.331 moles

Mass of unreacted sodium metal = 0.331 moles × 23 gmol-1= 7.613 g

If 1 mole of HNO3 yields 1 mole of hydrogen gas

9×10^-3 moles of HNO3 yields 9×10^-3 moles of hydrogen gas.

1 mole of hydrogen gas occupies 22.4 L

9×10^-3 moles of hydrogen gas will occupy 9×10^-3 moles × 22.4/1 = 0.20 L

3 0

3 years ago

When an egg is placed in a hot frying pan, the yolk and white become firmer and more opaque, and steam is emitted.

hodyreva [135]

Answer:

Formation of Gas, Color Change, and Absorption of heat is the Answer.

Explanation:

I just took the test.

5 0

3 years ago

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Two or more elements chemically combined make a?

In-s [12.5K]

Tqo or more elements combine to make a compound.

5 0

3 years ago