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2 years ago

A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a

Chemistry

1 answer:

san4es73 [151]2 years ago

3 0

Answer:

A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.

Explanation:

A simple substitution reaction or simple displacement reaction, called single-displacement reaction, is a reaction in which an element of a compound is substituted by another element involved in the reaction. The starting materials are always pure elements and an aqueous compound. And a new pure aqueous compound and a different pure element are generated as products. The general form of a simple substitution reaction is:

AB + C → A +BC

where C and A are pure elements; C replaces A within compound AB to form a new co, placed CB and elementary A.

So, in a Single replacement reaction an uncombined element replaces an element.

<u><em>A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.</em></u>

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4 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a

adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k

V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

$\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{10.1*746*273}{760*296}$

V2=9.1436cm³

6 0

3 years ago