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2 years ago

PLEASE SOMEBODY HELP PLEASE ! I WILL MARK BRAINLY I PROMISE !

Chemistry

1 answer:

IgorLugansk [536]2 years ago

5 0

A. your reactants are on the left of the arrow.
B. your products are on the left of the arrow.
C.its read as a yeild sign.
D.the s means it's a soild and the g means it's a gas.

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Which of these has the most covalent character? NaF Caf2 Bef2 CsF SrF2

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Bef2 is the compound that has the most covalent character. Whichever element has the greatest electronegativity has the least difference and most covalent character.

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3 years ago

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How many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?

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P₄ + 10Cl₂ ---> 4PCl₅
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3 years ago

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3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

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2 years ago