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3 years ago

A 100.5 ml intraveneous (iv) solution contains 5.10 g glucose (c6h12o6). what is the molarity of this solution?

1 answer:

8 0

Answer is: the molarity of this glucose solution is 0.278 M.
m(C₆H₁₂O₆) = 5.10 g.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆) .
n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆) = 0.028 mol.
V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆) = 0.278 mol/L.

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An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical

sladkih [1.3K]

Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help

7 0

3 years ago

Read 2 more answers

For the reaction 2 HCl + Ca ( OH ) 2 ⟶ CaCl 2 + 2 H 2 O how many grams of calcium hydroxide, Ca(OH)2, are needed to react comple

kari74 [83]

Answer: 65.38g of Ca(OH)2 is needed

Explanation:

From The equation of reaction

2 HCl + Ca ( OH ) 2 ⟶ CaCl 2 + 2 H 2 O

NB: Molar mass of HCl= 1+35.5=36.5

Ca(OH)2= 74

From The stoichiometric equation

2mol of HCl(36.5×2=73) require 1mol of Ca(OH)2 (74g)

Hence 64.5g of HCl will require 64.5×74/73= 65.38g of Ca(OH)2

5 0

2 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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5 0

2 years ago

The theoretical yield is the amount of product actually produced. a. True b. False

jolli1 [7]

Answer:

a. True

Explanation:

Theoretical yield is the amount of product that could be obtained if a chemical reaction has 100% efficiency.

Hope it helps...

6 0

3 years ago

Read 2 more answers

2. if 0.20 m fe3 had been used instead of 0.020 m fe3 , how would the numerical value of the rate constant and the activation en

dezoksy [38]

the calculated value is Ea is 18.2 KJ and A is 12.27.

According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.

At 500K, K=0.02s−1

At 700K, k=0.07s −1

The Arrhenius equation can be used to calculate Ea and A.

RT=k=Ae Ea

lnk=lnA+(RT−Ea)

At 500 K,

ln0.02=lnA+500R−Ea

500R Ea (1) At 700K lnA=ln (0.02) + 500R

lnA = ln (0.07) + 700REa (2)

Adding (1) to (2)

700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.

=ln [0.02/0 .07]

Ea= 2/35×100×8.314×1.2528

Ea =18227.6J

Ea =18.2KJ

Changing the value of E an in (1),

lnA=0.02) + 500×8.314/18227.6

= (−3.9120) +4.3848

lnA=0.4728

logA=1.0889

A=antilog (1.0889)

A=12.27

Consequently, Ea is 18.2 KJ and A is 12.27.

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5 0

1 year ago