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Zigmanuir [339]

2 years ago

14

WILLL GIVE BRAINLIEST IF YOU ANSWERRR PLEASEEEE IM LITERALLY BEGGING YOU I PUT THIS QUESTION IN SOO MANY TIMES I HAVEN'T GOTTEN

AN ANSWER PLEASEEE HELPPP

1 answer:

tino4ka555 [31]2 years ago

6 0

Answer:

The answer is 375.54 g of AgBr

Explanation:

Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)

Mass = 2M x 1L x 187.77 g/mol

Mass = 375.54g

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4. Calculate the final concentration if water is added to 1.5 L of a 12 M

konstantin123 [22]

Answer:

6M

Explanation:

(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted doln

Molarity dilute soln = [(M x V)conc/V (dilute)] = 1.5L x 12M / 3.0L = 6M final dilute soln

6 0

3 years ago

If a solution has a concentration of 12 M when the volume is 150 mL. What is the new concentration if you increase the volume to

swat32

Answer:

The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only ...

Explanation:

hope it helps you

6 0

2 years ago

Which of the following contains the most atoms?

aksik [14]

Explanation:

9.0122 g be Maybe, sorry i don't think so

8 0

3 years ago

Read 2 more answers

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago

If 9.6 X 1021 molecules of hydrogen are reacted with excess nitrogen, how many liters of ammonia can be produced at STP?

diamong [38]

The volume of NH₃ produced at STP : 0.237 L

<h3>Further explanation</h3>

Reaction

N₂ + 3H₂ → 2NH₃

1 mol = 6.02 x 10²³ particles

9.6 X 10²¹ molecules of Hydrogen, mol :

$\tt \dfrac{9.61\times 10^{21}}{6.02\times 10^{23}}=1.596\times 10^{-2}~moles$

mol H₂ : mol NH₃ = 3 : 2

mol NH₃ :

$\tt \dfrac{2}{3}\times 1.596\times 10^{-2}=0.0106$

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). <em>At STP, Vm is 22.4 liters/mol.</em>

The volume of NH₃ :

$\tt 0.0106\times 22.4=0.237~L$

<em />

8 0

2 years ago