I'm having with chemistry assignments I need a help

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iVinArrow [24]

Answer:

When these materials end up in the ocean, they support the growth and well-being of marine life.

Explanation:

This is the incorrect statement

8 0

3 years ago

Initial mass of triphenyl methanol in g = 0.220g Initial volume of 33% HBr solution in ml = 0.60 ml Find mas of triphenyl bromid

tensa zangetsu [6.8K]

Answer:

0.792g of triphenyl bromide are produced.

Explanation:

The reaction of triphenyl methanol with HBr is:

triphenyl methanol + HBr → Triphenyl bromide.

Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.

To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first limiting reactant:

Moles triphenyl methanol (Molar mass: 260.33g/mol) =

0.220g × (1mol / 260.33g) = 8.45x10⁻³ moles Triphenyl methanol

Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =

0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = 2.45x10⁻³ moles HBr

As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is limiting reactant.

As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = 2.45x10⁻³ moles

As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:

2.45x10⁻³ moles × (323.2g / mol) =

<h3>0.792g of triphenyl bromide are produced.</h3>

5 0

4 years ago

Please answer cansend to me

Ahat [919]

The answer is thank u for the points!

8 0

3 years ago

Read 2 more answers

The Ka of hydrazoic acid (HN3) is 1.9 × 10-5 at 25.0 °C. What is the pH of a 0.40 M aqueous solution of HN3?

siniylev [52]

Answer:

The pH of the solution is 2.56.

Explanation:

Given :

Concentration (c) = 0.40 M

Acid dissociation constant = $K_a=1.9\times 10^{-5}$

The equilibrium reaction for dissociation of $HN_3$ (weak acid) is,

$HN_3+H_2O\rightleftharpoons N_3^{-}+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

Dissociation constant is given as:

$K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

$1.9\times 10^{-5}=\frac{(0.40\alpha)(0.40\alpha)}{0.40(1-\alpha)}$

By solving the terms, we get value of $(\alpha)}$

$\alpha=0.00686832$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.4\times 0.00686832=0.002747 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.002747)$

$pH=2.56$

Therefore, the pH of the solution is, 2.56.

8 0

4 years ago

In the following chemical reaction, which compound is a precipitate? NaCl(aq) + AgNO3(aq) ⟶ NaNO3(aq) + AgCl (s) NaCl AgNO3 NaNO

makvit [3.9K]

Answer: AgCl

Explanation:

When the reaction takes place in which the sodium chloride is reacted with silver nitrate then the products formed are sodium nitrate and silver chloride.

NaCl(aq) + AgNO3(aq) ⟶ NaNO3(aq) + AgCl (s)

This reaction is known as precipitation reaction in which the white precipitate of silver chloride is formed.

This precipitate is insoluble and separates out of the solution.

So, the correction answer is AgCl(silver chloride).

4 0

4 years ago