What is the cornaviris?

Use the pop-up menus to label the following statements as either True of False.

goldenfox [79]

Answer:

1 True

2. True

3. True

4. True

5. False

Explanation:

1. Hypochlorous acid (HOCl) is a weak acid that forms when chlorine dissolves in water, and is partially dissociated, forming hypochlorite. HOCl is considered a strong oxidizing agent responsible for the killing action of germs in chlorine solutions.

2. Acetic acid is also known as ethanoic acid. It has a distinctive sour taste and pungent smell, and is generally considered a weak acid. in a concentrated form, acetic acid can be corrosive.

3/4 . Methylene chloride is an organochloride compound with the formula CH2Cl2. It is a colorless, volatile liquid with a moderately sweet aroma is widely used as a solvent. It is normally stable, non-flammable and non-explosive when mixed with air at temperatures below 100 °C, but its vapor is flammable only when present in relatively high concentrations of about 14% to 22% in air. Research shows that it is very much possibly carcinogenic, and it has been linked to cancer of the lungs, liver, pancreas, breast cancer and salivary gland cancer in laboratory animals.

5. Laboratory and medical waste are not to be disposed in the trash with regular wastes, as they are potentially toxic. Special disposal methods and agencies exists to take away such wastes.

6 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

Weather is not the same as climate. Which claim

Natalka [10]

C ........................

6 0

3 years ago

Read 2 more answers

Phosphoric acid is a triprotic acid with the following pKa values:

lisabon 2012 [21]

Answer:

Mass NaH₂PO₄ = 1.920 g

Mass Na₂HPO₄ = 4.827 g

Explanation:

For a buffer solution we know its pH can be calculated from the Henderson-Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] and [HA] are the concentrations of the weak acid and its conjugate base in the buffer.

We want to prepare a buffer at pH 7.540 so we have chosen salts NaH₂PO₄ and Na₂HPO₄ as the weak acid and conjugate base respectively.

To calculate the mass of these salts we need to compute their ratio in the Henderson- Hasselbach equation .

Now since we are asked to determine the masses of NaH₂PO₄ and Na₂HPO₄ and we know we want to prepare 1.000 L of a 0.05 M phosphate buffer, we can setup a system of 2 equations with two unknowns from the ratio mentioned above:

pH = pKa + log [A⁻]/[HA]

7.540 = 7.198 + log[HPO₄²⁻] / [H₂PO₄ ⁻]

0.342 = log[HPO₄²⁻] / [H₂PO₄ ⁻]

taking inverse log function to both sides of this equation:

2.198 = [HPO₄²⁻] / [H₂PO₄ ⁻]

but this is also equivalent to

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻ (M = mol/V)

We also know that in 1 liter of 0.05 M phosphate, we have 0.05 total mol HPO₄²⁻ and H₂PO₄⁻ , thus

mol HPO₄²⁻ + mol H₂PO₄⁻ = 0.05 mol

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻

solving this system of equations calling x = mol HPO₄²⁻ and y = mol H₂PO₄⁻ , we have:

2.198 = x /y ⇒ x = 2.198y

x + y = 0.05

2.198y + y = 0.05

3.198 y = 0.05 ⇒ y = 0.05 / 3.198 = 0.016

x = 0.05 - 0.016 = 0.034

and the masses can be calculated from the molar masses ( 141.96 g/mol Na₂HPO₄ and 119.98 g/mol NaH₂PO₄

mol HPO₄²⁻ = 0.034 mol x 141.96 g/mol = 4.827 g

mol H₂PO₄⁻ = 0.016 mol x 119.98 g/mol = 1.920 g

6 0

3 years ago

4. Substances that existed before mixing-still exist-after substances-are-mixed----

pantera1 [17]

False; Because when 2 substances are mixed another substance is created.

5 0

3 years ago