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3 years ago

Plzzzzzz help me I will mark u brainiest

Chemistry

1 answer:

Monica [59]3 years ago

3 0

Brainless answer:

I think it is ionic.. I am not sure

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Draw the skeletal structure for the monohalogenated alkene formed when trans-1-chloro-4-iodocyclohexane is reacted with 1 equiva

Ksivusya [100]

Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture

5 0

3 years ago

Help me with this plz

Roman55 [17]

Answer:

EVEN DOE IT TOO LATE, ITS B

Explanation:

8 0

2 years ago

If a stock solution of 20.0 g of sugar per 100. ml of solution is available, how much of this stock is needed to prepare 10.0 ml

Vladimir79 [104]

You must use 2.50 mL of the concentrated solution to make 10.0 mL of the dilute solution.

We can use the dilution formula

V_1C_1 = V_2C_2

where

V represents the volumes and

C represents the concentrations

We can rearrange the formula to get

V_2 = V_1 × (C_1/C_2)

V_1 = 10.0 mL; C_1 = 5.00 g/100. mL

V_2 = ?; ____C_2 = 20.0 g/100. mL

∴ V_2 = 10.0 mL × [(5.00 g/100. mL)/(20.0 g/100. mL)] = 10.0 mL × 0.250

= 2.50 mL

3 0

3 years ago

Calculate the ph of a 0.10 m solution of hypochlorous acid, hocl. ka of hocl is 3.5×10−8 at 25 ∘c. express your answer numerical

Viktor [21]

Answer:

4.23.

Explanation:

∵ pH = - log[H⁺].

For weak acids:

∵ [H⁺] = √(ka)(c).

∴ [H⁺] = √(3.5 × 10⁻⁸)(0.10 M) = 5.92 x 10⁻⁵.

∴ pH = - log[H⁺] = - log(5.92 x 10⁻⁵) = 4.2279 ≅ 4.23.

3 0

4 years ago

¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?

Archy [21]

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent$

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

5 0

3 years ago