A certain element has 12 neutrons and 12 protons. Assume that this atom has an

David walks 10 Km North. He turns East and walks 10 more Km. What is his distance

BigorU [14]

Answer:

20 km

Explanation:

he walks 10 km + another 10 km so 20 :)

3 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

Calculate the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remain

Reil [10]

The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:

3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂

This is a redox reaction, which <u>is a chemical reaction in which one or more electrons are transferred between the reagents</u>, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.

<u>The mass of chromium metal produced in the above reaction will be,</u>

425.0 mL x $\frac{1 L}{1000 mL}$ x $\frac{0.25 mol Cr(NO_{3})_{3} }{1 L}$ x $\frac{2 mol Cr }{2 mol Cr(NO_{3})_{3} }$ x $\frac{51.9961 g Cr}{1 mol Cr}$ = 5.52 g

So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.

6 0

3 years ago

The movement of one body around another is

Morgarella [4.7K]

Answer:

rotation

Explanation:

8 0

3 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in

ElenaW [278]

Answer: