All categories

4 years ago

A reaction produces 0.863 moles of H2O. How many molecules are produced?

Chemistry

2 answers:

Stolb23 [73]4 years ago

8 0

Number of molecules = moles x 6.02 x 10^23 = 0.863 x 6.02 x 10^23
= 5.19 x 10^23 molecules.

balu736 [363]4 years ago

4 0

To convert from moles to molecules, you multiply by Avogadro's constant.

0.863 moles H₂O * (6.022*10^23 molecules/1 mole) = 5.20*10^23 molecules H₂O

5.20*10^23 molecules of H₂O are produced.

You might be interested in

2 X(9) + Y (9) = 2XY (9)

11Alexandr11 [23.1K]

Answer:

The answer is D. The rate of the forward reaction is greater than the rate of the reverse reaction.

Explanation:

The forward reaction rate was greater than the reverse reaction rate because more product is present at time 2. This is the answer from ap classroom.

8 0

3 years ago

NEED HELP!!! PLEASE, AND THANK YOU!! ^^ <3

anygoal [31]

Answer:

a) how to prepare a verbal report

Explanation:

8 0

3 years ago

Read 2 more answers

C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????

Yanka [14]

Answer:

Rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$

Explanation:

According to Arrhenius equation for a reaction-

$ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

where $k_{2}$ and $k_{1}$ are rate constants of reaction at $T_{2}$ and $T_{1}$ temperatures (in kelvin) respectively.

$E_{a}$ is activation energy of reaction.

Here $T_{1}$ = 600 K , $k_{1}$ = $6.1\times 10^{-8}s^{-1}$

$T_{2}$ = 725 K, $E_{a}$ = 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

$ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})$

So, $k_{2}$ = $5.2\times 10^{-4}s^{-1}$

Hence rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$

7 0

4 years ago

Whichof the following is another name for science

snow_lady [41]

Field or discipline
hope it helped

8 0

3 years ago

1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many

UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0

3 years ago