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Marina86 [1]
3 years ago
9

The following questions involve the quantities of some materials you will be using in the calorimetry lab. Keep a copy of your a

nswers, so you will not have to recalculate them in the lab. (a) How many moles of NH3 will you be using in Part C of the calorimetry experiment?
Chemistry
1 answer:
Dovator [93]3 years ago
4 0

Answer:

0.075 moles

Explanation:

 Molarity = Mass/ Molar Mass*1000/Volume(ml)

Mole = Mass/ Molar mass

Given

  Molarity = 1.50 moles/ml

  Volume = 50 ml

∴ moles = Molarity*Volume/1000

              =1.50 *50  / 1000

              =0.075 moles

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melamori03 [73]

Answer:

Explanation:Reaction NaHCO3 + HCl ⇒ NaCl + H2O + CO2

Molar mass M = (22.99+1.008+ 12.01+3·16 ) g/mol. Calculate

Amount of substance n =m/M, n(NaCl) is equal.

M(NaCl) = 58.44 g/mol and mass m= n·M

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3 years ago
WILL NAME BRAINLIEST
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sulfur combines with hydrogen to form hydrogen sulfide H2 S by single covalent bond, while it form with potassium by ionic bond​
Alex

Answer: Ions may be defined as the element that contains either positive or negative charge over them. Two types of ions are cations and anions. The outermost electrons are involved in the formation of ions.

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8 0
3 years ago
Read 2 more answers
A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v
miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution C_1=?

Volume of the sport drink before dilution V_1=5.4 mL

Concentration of red dye in sport drink after dilution C_2=18.1 ppm

Volume of the sport drink after dilution V_2=25.0 mL

C_1V_1=C_2V_2( dilution )

C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}

C_1=83.80 ppm\approx 84.0ppm

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

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3 years ago
Calculate the mole fraction of benzene in a mixture that contains 45.2 g of benzene and 115.8 g of toluene.
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molecular weight of toluene = 92 g/mol  

C6 = 12 x 6 = 72  

H5 = 1 x 5 =5  

CH3 = 12 + 1 x 3 =15  

molecular weight of benzene = 78 g/mol  

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H6 = 1 x 6 = 6  

moles toluene:  

115.8 / (92 g/mol) = 1.25 mol  

moles benzene:  

45.2/ (78 g/mol) = 0.57mol  

total moles: 1.25  + 0.57= 1.82 moles  

mole fraction of toluene:  

1.25 mol  / 1.82 moles  = 0.68



6 0
4 years ago
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