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3 years ago

How many grams of ammonia produced from 1000 grams of N2?

Chemistry

1 answer:

sleet_krkn [62]3 years ago

5 0

Answer:

N2 + 3H2 ———> 2NH3

As we know 1000 grams ammonia is 58.82 moles so according to unitary method,

2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.

No. Of moles = given mass/molar mass

Implies that

Mass of nitrogen required = 29.41*28 = 823.48 grams.

Explanation:

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Please help fast, I will give brainliest.

frez [133]

Answer:

The reaction is favorable at all temperatures

Explanation:

Since G = H - TS, -H and +S would result in G = -H -TS, which will always be negative.

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3 years ago

A flask contains 5.0g of neon gas at a temperature of 35°C. The pressure gauge indicates 0.37atm inside the flask. What is the v

DaniilM [7]

Answer:

The volume of the neon gas is 17.07 L.

Explanation:

The ideal gas equation describes the relationship among the four variables P, V, T, and n. An ideal gas<u> is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation</u>.

In order to calculate the volume, first we need to convert the grams of neon to moles:

28.18 g Ne ------ 1 mol Ne

5.0 g Ne---------- <u>x= 0.25 mol Ne</u>

Now, using the ideal gas equation we calculate the volume:

$pV= nRT$

$V= nRT ÷ p$

$V= (0.25 mol × 0.082 L.atm/mol.K × 308.15 K ÷ 0.37 atm$

V= 17.07 L

6 0

3 years ago

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro

deff fn [24]

Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Mass of the water = 25.0 grams

Initial temperature of the metal = 90.0 °C

Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

Qmetal = -Qwater

m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒ with mass of metal = 12.0 grams

⇒ with c(metal) = TO BE DETERMINED

⇒ with ΔT(metal) = T2 - T1 = 25.0°C - 90.0 °C = -65.0 °C

⇒ with mass of water = 25.0 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C

-780.0 * c(metal) = -2615 ( 2.6*10^3 with sig figs)

c(metal) = 0.335 (=0.34 with sig figs)

The specific heat of the metal is 0.34 J/g°C

3 0

3 years ago

Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

4 0

3 years ago

A farmer wants to expand his fields, but only has hills left. He is worried about water erosion because of the rate at which wat

Aleks04 [339]

Answer:

Reshape the hills with terraces

Explanation:

It is the best way of preventing water eroding when it comes of doing agriculture on hills

7 0

3 years ago