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Vinvika [58]
2 years ago
10

A solution of water (kb=0.512 ?c/m) and glucose boils at 103.06 ?c. what is the molal concentration of glucose in this solution?

assume that the boiling point of pure water is 100.00 ?c.
Chemistry
2 answers:
AfilCa [17]2 years ago
6 0

Answer:

 

100.26°C

Explanation:

nirvana33 [79]2 years ago
4 0
<span>When a solute is added to a solvent, some properties are affected and these set of properties are called colligative properties. The properties depend on the amount of solute dissolved in a solvent. For this we use the boiling point elevation of the solution. We do as follows:

</span>ΔT(boiling point) = 103.06 °C - 100.0 °C= 3.06 °C<span>
ΔT(boiling point)  = (Kb)m
m = </span>ΔT(boiling point) / Kb<span>
m =  3.06 °C / 0.512 °C kg / mol
m = 5.98 mol / kg

</span>
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Which of these is a difference between a single skin cell and the skin tissue that covers an entire hand?
fgiga [73]

Answer:

c

Explanation:

the cells grow more than the tissue

7 0
2 years ago
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A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?
Dominik [7]

Answer:

The specific heat of the metal is 0.466 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

  • Q= 2330 J
  • c= ?
  • m= 25 g
  • ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

c=\frac{2330 J}{25 g* 200 C}

c=0.466 \frac{J}{g*C}

<u><em>The specific heat of the metal is 0.466 </em></u>\frac{J}{g*C}<u><em></em></u>

6 0
2 years ago
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Water treatment plants commonly use chlorination to destroy bacteria. a byproduct is chloroform (chcl3), a suspected carcinogen
antiseptic1488 [7]
<span>100. ppb of chcl3 in drinking water means  100 g of CHCl3 in 1,000,0000,000 g of water

Molarity, M

M = number of moles of solute / volume of solution in liters

number of moles of solute = mass of CHCl3 / molar mass of CHCl3

molar mass of CHCl3 = 119.37 g/mol

number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol

using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters

M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer

Molality, m

m = number of moles of solute / kg of solvent

number of moles of solute = 0.838

kg of solvent = kg of water = 1,000,000 kg

m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer

mole fraction of solute, X solute

X solute = number of moles of solute / number of moles of solution

number of moles of solute = 0.838

number of moles of solution = number of moles of solute + number of moles of solvent

number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles

number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles

X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer

mass percent, %

% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =

% = 10 ^ - 6 % <------- answer
</span>
7 0
2 years ago
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A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

n = 0.146 moles

for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
2 years ago
The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 2.69 × 1012 at 25°c. calculate e o for a cell m
Mariulka [41]
Sr(s)+Mg²+(aq)→Sr²+(aq)+Mg(s)
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
=-(-70922.3)/2×96500)
=0.367v.
8 0
2 years ago
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