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worty [1.4K]
3 years ago
7

An industrial vat contains 650 grams of solid lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with exc

ess hydrochloric acid. This is the equation of the reaction: 2HCl + Pb(NO3)2 → 2HNO3 + PbCl2. What is the percent yield of lead(II) chloride? The percent yield of lead chloride is %
Chemistry
2 answers:
aleksandrvk [35]3 years ago
6 0

Answer:

             88.98 %

Solution:

The Balance Chemical Equation is as follow,

                                   2 HCl + Pb(NO₃)₂     →    2 HNO₃ + PbCl₂

According to equation,

           331.2 g (1 mole) Pb(NO₃)₂ produces  =  278.1 g (1 mole) PbCl₂

So,

                  870 g of Pb(NO₃)₂ will produce  =  X g of PbCl₂

Solving for X,

                      X  =  (870 g × 278.1 g) ÷ 331.2 g

                      X  =  730.5 g of PbCl₂

Therefore,

                Theoretical Yield =  730.5 g

Also as given,

                 Actual Yield =  650 g

So using following formula for percentage yield,

                         %age Yield  =  (Actual Yield / Theoretical Yield) × 100

Putting values,

                         %age Yield  =  (650 g / 730.5 g) × 100

                         %age Yield  =  88.98 %

Masteriza [31]3 years ago
3 0

Answer:

89%

Explanation:

Round it up

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cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
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Explanation: 
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This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
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Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
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To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
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We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
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Then we write:  "  * 10^ "
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   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
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So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
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Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
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Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
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{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
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The answer is:  " 1.75 * 10^(-10)   m " .
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EastWind [94]

Answer:

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Explanation:

Given parameters:

Initial volume  = 34.3mL    = 0.0343dm³

Initial concentration  = 1.72mM   = 1.72 x 10⁻³moldm⁻³

Final concentration  = 1.00mM = 1 x 10⁻³ moldm⁻³

Unknown:

Final volume  =?

Solution:

Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.

   Therefore;

              C₁V₁  =  C₂V₂

where C and V are concentration and 1 and 2 are initial and final states.

        now input the variables;

                      1.72 x 10⁻³ x  0.0343 = 1 x 10⁻³  x V₂

                        V₂ = 0.0589dm³ = 58.9mL

         

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zheka24 [161]

Answer:

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4 0
2 years ago
Read 2 more answers
If you have 1.1 moles of magnesium nitrate then how many grams is that
marysya [2.9K]

Answer: 162.8 grams

Explanation:

Magnesium nitrate has a chemical formula of Mg(NO3)2.

Given that:

Number of moles of Mg(NO3)2 = 1.1 moles

Mass in grams of Mg(NO3)2 = ?

For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g

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1.1 moles = Mass / 148g/mol

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Mass = 162.8 grams

Thus, there are 162.8 grams of magnesium nitrate.

3 0
3 years ago
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