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stepan [7]
3 years ago
15

HELP!!!

Chemistry
1 answer:
ruslelena [56]3 years ago
7 0
A positively charged ion is attracted to a negatively charged ion. ... represents the simplest ratio of the ions involved in the ionic compound. Oxyanion. is a polyatomic ion composed of an element and one or more oxygen atoms.
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All halogens are stable so they have 8 electrons in their last shell
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Properties of iron, barium, and phosphorus. Explain why nails are made of iron but they are never made of barium or phosphorus.
bearhunter [10]

Nails are made up of iron not of phosphorus and barium.  

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3 0
3 years ago
All of the following institutions awards certificates except: a.) Community college b.) liberal arts college c. Junior college d
Serjik [45]
Except B, Liberal Arts College
3 0
3 years ago
Read 2 more answers
A mineral consisted of 29.4% calcium, 23.5% sulfur and 47.1% oxygen. What is the empirical formula?
inessss [21]
Step  one  calculate  the  moles  of  each  element
that  is  moles= %  composition/molar  mass
molar  mass  of    Ca  =  40g/mol,     S=  32  g/mol ,   O=  16 g/mol

moles  of      Ca = 29.4 /40g/mol=0.735 moles,      S= 23.5/32  =0.734 moles,  O= 47.1/16= 2.94   moles

calculate  the mole  ratio  by  dividing   each  mole with  smallest  mole   that   is  0.734
Ca=  0.735/0.734= 1,      S=  0.734/0.734 =1,   O  = 2.94/  0.734= 4
therefore  the  emipical  formula  =  CaSO4
5 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
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