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Fiesta28 [93]
2 years ago
15

a metal worker uses a cutting torch that operates by reacting acetylene gas, C2H2(g), with oxygen gas, O2(g), as shown in the un

balanced equation below: C2H2(g)=O2(g)= CO2(g) = H2O(g) = heat determine the mass of 25 moles of acetylene (gram-formula mass = 26 g/mol
Chemistry
2 answers:
harkovskaia [24]2 years ago
8 0

Answer:

650 grams

Explanation:

Given that acetylene gas reacts with oxygen to produce caobon dioxide, water and heat and the unbalnced equation is

C_2H_2(g)+O_2(g) \rightarrow CO_2(g) + H_2O(g)+ heat

Gram-formula mass of C_2H_2= 26 g/mol

So, mass of 1 mole of acelylene is 26 grams

Therefore, mass of 25 moles of acelylene=25x26=650 grams

Hence, the mass of 25 moles of acelylene is 650 grams

erastova [34]2 years ago
6 0

Answer:

650

Explanation:

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goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

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How many moles of O2 would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 73.3% yi
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Mmmmmmmmmmmmmmm

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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
mars1129 [50]

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0
2 years ago
Which best represents the reaction of calcium and zinc carbonate (ZnCO3) to form calcium carbonate (CaCO3) and zinc? Ca → ZnCO3
artcher [175]
Ca + ZnCO3 → CaCO3 + Zn
3 0
2 years ago
Read 2 more answers
When 200g of AgNO3 solution mixes with 150 g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution ri
gizmo_the_mogwai [7]

Answer:

\Delta H=1962.3J

Explanation:

Hello,

In this case, we can compute the change in the solution enthalpy by using the following formula:

\Delta H=mC\Delta T

Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:

\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J

It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.

Best regards.

8 0
2 years ago
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