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3 years ago

Select the most likely product for this reaction:

Chemistry

2 answers:

ki77a [65]3 years ago

7 0

Answer:

2) LiNO3(aq) + H2O(l)

Explanation:

Advocard [28]3 years ago

3 0

Answer:

The answer is B- LiNO3(aq)+H2O(I)

Explanation:

I just did the assignment also UNUS ANNUS

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What does the term division of labor mean as it relates to cells?

lidiya [134]

Answer:

what each cells job is divided

Explanation:

7 0

3 years ago

How can so many different substances in the world be made of so few elements?

Mrrafil [7]

Matematically speaking, maybe because:
The number of substances = number of elements + number of different combinations of those elements

7 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

Which of the following statements is true of enzymes A. They're unusable after a reaction B. They don't provide an active site f

Mnenie [13.5K]

D. They act on a specific type of substrate in a reaction

Hope this helps!

3 0

3 years ago

Read 2 more answers

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago