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ANEK [815]
3 years ago
8

Given the reaction: 4 NH3(g) + 5 O2(g) → 4 NO (g) + 6 H2O When 1.20 mole of ammonia reacts, how many moles of water are produced

?
Chemistry
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

When 1.20 mole of ammonia reacts, 1.8 moles of water are produced.

Explanation:

The balanced reaction is:

4 NH₃(g) + 5 O₂(g) → 4 NO (g) + 6 H₂O

By stoichiometry of the reaction, the following amounts of moles participate in the reaction:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O:  6 moles

Then you can apply the following rule of three: if by stoichiometry 4 moles of ammonia produce 6 moles of water, 1.2 moles of ammonia will produce how many moles of water?

moles of water=\frac{1.2 moles of ammonia*6 moles of water}{4 moles of ammonia}

moles of water= 1.8 moles

<u><em>When 1.20 mole of ammonia reacts, 1.8 moles of water are produced.</em></u>

<u><em></em></u>

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2 years ago
Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.
antiseptic1488 [7]

Answer: 45.983 g CBr₄

Explanation:

To convert from moles to grams, you know that we will need molar mass and Avogadro's number.

Avogadro's number: 6.022×10²³ molecules/mol

Molar mass: 331.627 grams/mol

Now that we have what we need, you can use these to solve for grams. 8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g

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4 0
3 years ago
How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C
yaroslaw [1]

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

  • c= 4.184 \frac{J}{g*C}
  • m=21 g
  • ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 \frac{J}{g*C} * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

<u><em> 527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>

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