All categories

2 years ago

What are the products of the following equation?

Chemistry

1 answer:

Artyom0805 [142]2 years ago

5 0

The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4

You might be interested in

fish give off the compound ammonia which has a pH above 7 to which class of compounds does ammonia belong

jasenka [17]

Answer:
The compound ammonia given by fish is alkaline

Explanation:
We can classify elements/compounds based on their pH values into three types:
acids: these are compounds having pH value lower than 7
neutral: these are compounds having pH value equal to 7
alkalies: these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)

5 0

3 years ago

PLEASE HELP!! this is on USAtestprep a) b) c) d)

Rudik [331]

The answer would be A from what I know

3 0

3 years ago

The student adds 0.0010 mol of NaOH(s) to solution Y, and adds 0.0010 mol of NaOH(s) to solution Z. Assume that the volume of ea

julia-pushkina [17]

Answer:

The answer is in the explanation.

Explanation:

A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Buffers are able to avoid the pH change of a solution when strong acid or bases are added (As NaOH).

Based on the experiment, it is possible that the solution Z was a buffer and Y another kind of solution. For this reson, pH of the solution Y changes much more than the pH of solution Z changes despite the amount of NaOH added is the same in both solutions.

6 0

2 years ago

Codeine C18H21NO3 is a weak organic base. A 5.0 x 10^-3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for th

arsen [322]

Answer:

Kb = $4.45\times10^{-7}\ mol/L$

$p^{Kb}=6.35$

Explanation:

For a weak organic base, the formula to find $p^{OH}$ is given by:

$p^{OH}=p^{K_b}+\log c$

where c is the concentration of base.

Here c= $5\times10^{-3}\ M$

$p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05$

Substituting the above values in the formula,we get:

$p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L$

Hence:

Kb = $4.45\times10^{-7}\ mol/L$

$p^{Kb}=6.35$

7 0

2 years ago

The molar mass of the metal oxide M2O is 231.74 g/mol. What is the molar mass of the chloride of this metal?

Semenov [28]

The molar mass of the M2O = 2(molar mass of M) + molar mass of oxygen
Molar mass of oxygen = 16
Therefore:
231.74 = 2(molar mass of M) + 16
215.74 = 2*molar mass of M
molar mass of M = 107.87 grams

Now, from the given formula, we can notice that the metal M is monovalnet and we know that chlorine is monovalent as well.
Therefore, the compound formed will be: MCl
molar mass of chlorine = 35.5 grams
Therefore,
molar mass of MCl = 107.87 + 35.5 = 143.37 grams

5 0

3 years ago