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2 years ago

Please help I’ll mark you brainiest if it’s correct. There is 3 answers to this one!

Chemistry

1 answer:

Rama09 [41]2 years ago

7 0

Answer:

c,e,f

Explanation:

Judging by what i tried to find out about this reaction, it seems that won't take place, though it could be wrong.

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Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony t

Mrrafil [7]

Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-

3 0

2 years ago

How many moles are in 5g of Al? A.0.185 B.18.5 C.5.4 D.135

Lapatulllka [165]

I got 134.91 but if you round it you’ll get 135

3 0

3 years ago

Read 2 more answers

What is sin? can you explain me? anyone?l

REY [17]

Answer:

In a religious context, sin is an act of transgression against divine law. Each culture has its own interpretation of what it means to commit a sin.

Explanation:

3 0

3 years ago

Which description gives the steps for creating hydroelectric power

Inessa05 [86]

Answer:

The answer to your problem is

B) river water stored in a reservoir flows down a channel in a dam, turning a turbine as it passes. the turbine is connected to a generator that produces electricity, and then the water continues downstream

plz can i get brainliest

Explanation:

3 0

3 years ago

Read 2 more answers

How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride

Pepsi [2]

Answer:

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

Explanation:

Formula of Gold (III) chloride: $AuCl_{3}$

Avogadro Number : Number of particles present in one mole of a substance.

${N_{0}} =6.022 \times 10^{23}$

Using,

$n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}$

$n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}$

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

$6.022 \times 10^{23}$ = 1 mole of $AuCl_{3}$

In this Sample,

1 mole of $AuCl_{3}$ will give = 3 mole of Chlorine atoms

1 mole of Cl contain = $6.022 \times 10^{23}$

3 mole of Cl contain = $6.022 \times 10^{23}\times 3$

3 mole of Cl contain = $18.066 \times 10^{23}$

So,

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

8 0

3 years ago