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d1i1m1o1n [39]
3 years ago
8

What compound is SnBr4?

Chemistry
2 answers:
raketka [301]3 years ago
4 0
<span><span>SnBr4 is Bromide. Hope this helps!</span></span>
Maslowich3 years ago
3 0
The compound is Tin(IV) bromide
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Is burning of coal a reversible change???<br> If yes how?
koban [17]
I don't think so. No way that I know anyway. It it could be done then the need for more coal to be mined would have stopped hundreds of years ago. Once coal is burned, it forms water and carbon dioxide (essentially) with some sulfur oxides. 

How do you put that back together again. It's a little like humpty dumpty.
5 0
3 years ago
Read 2 more answers
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?
3241004551 [841]

Answer:  " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

<u>____________________________</u>

Explanation:

<u>____________________________</u>

<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

____________________________

The "balanced chemical equation" is:

____________________________

        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

_____________________________

<u>Note</u>: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

____________________________

So:  We call solve using a technique known as:  "dimensional analysis" :

____________________________

  0.500 mol AlCl₃ * (\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?

____________________________

<u>Note</u>:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

____________________________

 " \frac{(0.500 * 2 * 26.98)}{2}   g Al ["grams of aluminum"] ;

____________________________

<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

    = 13.49 g Al ;

         →  Round to 3 (Three) significant figures;

         →  Since:  "0.500" has 3 (Three) significant figures:

____________________________

   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

____________________________

 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0
3 years ago
Consider the addition of HBr to 2-pentene. Indicate the re
lana [24]

Answer:

This addition reaction yields 3-BromoPentane and 2-BromoPentane.

Explanation: The reaction is an addition reaction that follows the Markonikoff's principle engaging the electrophillic addition mechnism with electrophile having no lone pair so rearrangement of carbonation is possible. It yields two possible products.

5 0
3 years ago
Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H
lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

6 0
3 years ago
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