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2 years ago

5

Hi, im really confused can someone help

1 answer:

GenaCL600 [577]2 years ago

3 0

Lol I don’t know anything about this sorry

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A balanced chemical equation used to prepare ammonium carbonate, (nh4)2co3 , is: 2 n h 3 ( g ) + c o 2 ( g ) + h 2 o ( l ) ⟶ ( n

Tems11 [23]

Answer:

This a simple stoichiometry problem using the ideal gas law.

First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol

Now use a molar conversion using the balanced equation,

1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas

PV=nRT

V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L

7 0

3 years ago

The density of cadmium is 8.65 g/mL, what is the volume of a 33.4 g object made of pure cadmium?

Rzqust [24]

Answer:

Volume = 3.86 ml (Approx)

Explanation:

Given:

Density of cadmium = 8.65 g/ml

Mass of pure object = 33.4 g

Find:

Volume pure cadmium

Computation:

Volume = Mass / Density

Volume = 33.4 / 8.65

Volume = 3.86 ml (Approx)

6 0

3 years ago

9. A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent. A l

Arte-miy333 [17]

We have that the the liquid is

C_2H_5OH (ethanol
And at a condition of H_2SO4 as catalyst and temp 170

From the question we are told

A student wished to prepare <em>ethylene </em>gas by <em>dehydration </em>of ethanol at 140oC using sulfuric acid as the <em>dehydrating </em>agent.
A low-boiling liquid was obtained instead of ethylene.
What was the liquid, and how might the reaction conditions be changed to give ethylene

<h3>Ethylene formation</h3>

Generally the equation is

2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20

Therefore

with ethanol at 140oC

The product is diethyl ethen

The reaction at 170 ethylene will give

C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)

Therefore

The the liquid is

C_2H_5OH (ethanol

For more information on Ethylene visit

brainly.com/question/20117360

8 0

1 year ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

Measurements of the heights of various plants in an experiment are called

Bess [88]

Answer:

Option 1 Data

5 0

2 years ago

Read 2 more answers