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3 years ago

Predict what would happen if the ammonium system described in the experiment was heated.

1 answer:

7 0

Answer is: concentration of products increases (ammonia nad water).
Chemical reaction: heat + NH₄⁺ + OH⁻ ⇄ NH₃ + H₂O.
According to Le Chatelier's Principle, the position of equilibrium moves to counteract the change, because heat is increased, system consume that heat, so equilibrium is shifted to right, by decreasing concentration of reaactants and increasing concentration of product.

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Two (2) moles of an ideal gas are in a container at 200 kpa and a temperature of 300 k. The volume occupied by the gas in the co

k0ka [10]

The volume occupied by the gas in the container is 1 m³

Boyles law applies

P₁ V₁ = P₂ V₂

Where P₁ = 200kpa

P₂ = 300kpa

if its initial volume is 1.5

then,

P₁ V₁ = P₂ V₂

200 × 1.5 = 300 × V₂

V₂ = 200 × 1.5 / 300

= 1 m³

Hence the volume occupied by the gas container is 1 m³

Learn more about the Boyles law on

brainly.com/question/13759555

#SPJ4

4 0

2 years ago

PLEASE HELP ME ASAP I WILL MARK BRAINLYEST AND I WILL ALSO GIVE 5 EXTRA POINTS

sdas [7]

Answer:

Jack Beacuse...

Explanation:

the soil has water in it so it will be more than one matterial

5 0

2 years ago

Read 2 more answers

a chef is using honey in a recipe. The honey is too thick to pour out of the jar. How could the chef make the honey easier to po

Andrej [43]

Heat the jar of honey up enough to where you can pour it

5 0

3 years ago

Read 2 more answers

After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?

FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: Mg₃N₂

I hope it helps!

5 0

3 years ago