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baherus [9]
2 years ago
8

(Brainliest) Please Help! Please do not give me a random, gibberish answer or I will report you and give you a low rating theref

ore dropping your popularity. Also, if you don't know the answer you can put your guess in the comments. I will give a Brainliest and 50 points!
What mass of KBr is present in 75.0 mL of 0.250 M KBr solution?
Chemistry
1 answer:
steposvetlana [31]2 years ago
6 0

Answer:

Answer : The correct answer for mass of KBr = 2.53 g

Given :

Molarity of KBr solution = 0.85 M

Volume of KBr solution = 25 mL

Converting volume from mL to L ( 1 L = 1000 mL )

Volume of solution = 25 mL * \frac{1 L }{1000mL}Volumeofsolution=25mL∗

1000mL

1L

Volume of solution = 0.025 L

Mass of KBr = ?

Mass of KBr can be calculated using following steps :

1) To find mole of Kbr :

Mole of KBr can be calculated using molarity .

Molarity : It is defined as mole of solute present in volume of solution in Liter .

It uses unit as M or \frac{mol}{L}

L

mol

It can be expressed as :

Molarity = \frac{mol of solute (mol)}{volume of solution (L)}Molarity=

volumeofsolution(L)

molofsolute(mol)

Plugging value of molarity and volume

0.85 \frac{mol}{L} = \frac{mol of Kbr}{0.025 L}0.85

L

mol

=

0.025L

molofKbr

Multiplying both side by 0.025 L

0.85 \frac{mol}{L} * 0.025 L = \frac{mole of KBr}{0.025 L} * 0.025 L0.85

L

mol

∗0.025L=

0.025L

moleofKBr

∗0.025L

Mole of KBr = 0.02125

2) To find mass of Kbr :

Mass of Kbr can be calculated using mole . Mole can be expressed as :

Mole (mol) = \frac{mass (g) }{molar mass \frac{g}{mol} }Mole(mol)=

molarmass

mol

g

mass(g)

Mole of Kbr = 0.02125 mol

Molar mass of KBr = 119.00 \frac{g}{mol}

mol

g

Plugging values in mole formula

0.02125 mol = \frac{mass (g)}{119.00 \frac{g}{mol}}0.02125mol=

119.00

mol

g

mass(g)

Multiplying both side by 119.00 \frac{g}{mol}

mol

g

0.02125 mol * 119.00 \frac{g}{mol} = \frac{mass (g)}{119.00 \frac{g}{mol}} * 119.00\frac{g}{mol}0.02125mol∗119.00

mol

g

=

119.00

mol

g

mass(g)

∗119.00

mol

g

Mass of KBr = 2.53 g

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Answer:

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Explanation:

<em>"The excretory system is a passive biological system that removes excess, unnecessary materials from the body fluids of an organism, so as to help maintain internal chemical homeostasis and prevent damage to the body. The dual function of excretory systems is the elimination of the waste products of metabolism and to drain the body of used up and broken down components in a liquid and gaseous state"</em>

5 0
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determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)
sasho [114]

The  mass  of water formed  is


<u><em>calculation</em></u>

Use  the  ideal   gas  equation   to  calculate the  moles of  NH3  and O2

that  is  Pv= n RT

where;  P= pressure,  

V=  volume,

n = number  of  moles,

R=gas   constant  = 0.0821  l .atm/ mol.K

make n the formula of  the subject  by diving   both side  by  RT

n =  PV /RT

The   moles of NH3

n= (1.50 atm  x 12.5 L) /(  0.0821 L. atm /mol.k   x 298 K)  =0.766  moles

The  moles  of  O2

=(1.1 atm  x 18.9  L) /  (  0.0821 L. atm/ mol.k   x 323 K) = 0.784  moles


write the reaction  between  NH3  and  O2

4 NH3  + 5 O2  →4 No  +6H2O


from  equation above  0.766  moles of NH3  reacted to produce  

0.766 x 6/4 =1.149 moles of H2O


0.784  moles of O2   reacted to  produce  0.784  x 6/5=0.9408  moles  of H20


since  O2  is totally  consumed, O2  is the limiting  reagent  and therefore  the  moles of H2O  produced=  0.9408  moles


mass  of  H2O  = moles x molar mass

 from  periodic table the  molar mass  of H2O  =  (1 x2)+16= 18  g/mol

mass = 18 g/mol  x 0.9408  moles= 16.93  grams


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Which of the following can form a hydrogen bond with the HF molecule?
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NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

<h3>What is a limiting reagent?</h3>

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

2NO(g) +2H_2(g) → N_2 +2H_2O

Moles of nitrogen monoxide

Molecular weight: M_(_N_O_)=30g/mol

n_(_N_O_) =\frac{mass}{molar \;mass}

n_(_N_O_) =\frac{22.0}{30g/mol}

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Moles of hydrogen

Molecular weight: M_(_H_2_)=30g/mol

n_(_H_2_) =\frac{mass}{molar \;mass}

n_(_H_2_) =\frac{5.80g}{2g/mol}

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NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

n_(_N_2_) = (2.9 mol- 0.73 mol NO x \frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}) x \frac{2g \;of \;H_2}{mole \;of \;H_2}

n_(_N_2_) =4.34 g

Learn more about limiting reagents here:

brainly.com/question/26905271

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