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WARRIOR [948]
3 years ago
14

Solid aluminum reacts with solid lodine to. produce solid aluminum iodide.

Chemistry
1 answer:
Step2247 [10]3 years ago
6 0

Answer:

After the arrow, usually the right. Balance NO+O2--->NO2. 2NO+O2--->2NO2 ... What is the symbol for "solid"? (s). 3Mg3 (PO4)2.

Explanation:

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PLEASE HELP ME OUT!!
vesna_86 [32]

Answer:

hydrogen

Explanation:

7 0
3 years ago
10 To become positively charged, an atom must: A Galn a proton B Lose a proton C С. Galn an electron D Lose an electron } ​
Studentka2010 [4]

Answer:

D: lose an electron

Explanation:

when an atom loses an electron it's positively charged and when it gain an electron it is negatively charged

6 0
3 years ago
Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Asp
viva [34]

Answer:

325mg of Aspririn

Explanation:

First you should note the information that the problem gives you:

- The bottle of Aspirin has 5gr (grains)

- 1gr(grain) = 65mg (miligrams)

Also, the problem is asking about how many aspirin are in 5 gr (grains), so you should use a conversion factor, as follows:

-First you should put the quantity you need to convert:

5grAspirin

-Then you write the denominator of the conversion factor that must have the same units that you want to convert, in this case gr:

5grAspirin*\frac{}{1grAspirin}

-Then you write the numerator with the units that you want to obtain and the numerical equivalence between the units, in this case:

5grAspirin*\frac{65mgAspririn}{1grAspirin}

-Finally you multiply numerators and divide by denominators:

5grAspirin*\frac{65mgAspririn}{1grAspirin}=325mgAspririn

8 0
3 years ago
Give three example of univalent element​
Aloiza [94]

Answer:

Hydrogen and Chlorine

Explanation:

They are both an example in univalent atoms, because of their nature to form only one single bond.

I wasn't able to find another example, hope it helped! :)

4 0
2 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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