24. The only thing that Carbon, nitrogen and oxygen want is

Please explain how to do it as well!

kogti [31]

Answer:

a. $\bold{C_{6}H_{12}O_{6}\rightarrow 6CO_{2}+6H_{2}O}$

△H=−72 kcal

The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]

b.

$\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}$

The iron(III) ions and chloride ions remain aqueous and are spectator ions in a reaction that produces solid barium sulfate.

7 0

3 years ago

Read 2 more answers

An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm

frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

xA= 0.34, yA= 0.55

ii) 0.50 = 0.55 nv + 0.34 nL

Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0

3 years ago

How many uL are present in 250 mL of H20? (1 uL = 10^-6 Liters)

Vanyuwa [196]

Answer:

250000 μL

Explanation:

If 1 L = 1000 mL

Then X L = 250 mL

X = (1 × 250) / 1000 = 0.25 L

Now we can calculate the number of microliters (μL) in 0.25 L:

if 1 μL = 10⁻⁶ L

then X μL = 0.25 L

X = (1 × 0.25) / 10⁻⁶ =250000 μL

4 0

3 years ago

How many protons, neutrons, and electrons make up an ion of

yKpoI14uk [10]

Answer:

protons=15

electron=15

neutron=16

6 0

3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

7 0

3 years ago