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3 years ago

Can someone please explain this to me t,t

Chemistry

1 answer:

Oksana_A [137]3 years ago

8 0

Answer:

Theoretical 32.04 g

The precent yeild 87. 39 %

Explanation:

Mole of H2 = m/mw = 4/2 = 2 mol

From the equation

1 mol H2 --> 1/2 mol CH3OH

2 mol H2 -->??

So we have 1 mol Ch3oh

Mass of Ch3oh = mole *Mw = 1* 32.04

= 32.04g >>> the theoretically

For percent yeild = actual / theoretical * 100%

( 28.0/32.04) *100% =

87.93%

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

iVinArrow [24]

Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

i is the van´t Hoff factor

Kb the boiling constant which for water is 0.512 ºC/molal

m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb = Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

= 9273.25 g / 62.07 g/mol =149.4mol

molality solution = 149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb = Kbm = 0.512 ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

6 0

3 years ago

At a higher temperature, K = 1.8 × 105. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will

Soloha48 [4]

The question is incomplete, the complete question is;

1. Given: 2A(g) <-> 2B(g) + C(g). At a particular temperature, K = 1.6x104.

At a higher temperature, K = 1.8x10-5. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will...

cause [A] to increase

cause [B] to increase

gave no effect

cannot be determined

Answer:

cause [B] to increase

Explanation:

We have to have it behind our minds that; for an exothermic reaction, when the temperature is increased, the value of the equilibrium constant decreases. On the other hand, for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant.

In this particular case, increasing the temperature decreased the equilibrium constant from K = 1.6x104 to K = 1.8x10-5 meaning that the equilibrium constant decreased with increase in temperature. This means that the reaction is exothermic and a decrease in temperature will favour the forward reaction hence more B is produced.

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3 years ago

The model of the atom evolved over time as scientific investigations uncovered new properties and observations about the atom. D

wariber [46]

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4 0

3 years ago

Bromophenol blue is the indicator used in detecting the endpoint for the antacid analysis in this experiment. what is the expect

Paul [167]

Hello!

For the antacid analysis, the chemical reactions that occur in the titration are the following ones:

First, the antacid (composed of weak bases and carbonates) is completely neutralized by the H⁺ ions in the HCl

2HCl + CaCO₃ → CO₂ + H₂O + 2CaCl₂

HCl + OH⁻ → H₂O + Cl⁻

The titration reaction consists in titrating the excess H⁺ ions that are left in the solution, by the following reaction:

H⁺ + NaOH → H₂O + Na⁺

So, when the equivalence point is reached, the solution will go from acid to basic. As bromophenol blue is yellow in acidic solution and blue in basic solution, you'll expect the indicator to change from yellow to blue.

Have a nice day!

7 0

3 years ago

Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (

Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0

3 years ago