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3 years ago

Given the reaction: 4NH3 + 502 4NO + 6H2O

Chemistry

1 answer:

Ainat [17]3 years ago

8 0

Answer:

$\frac{4}{5}$ moles

Explanation:

The reaction equation is given as:

4NH₃ + 5O₂ → 4NO + 6H₂O

The number of moles of O₂ that completely reacted is given as 1 mole

To solve this problem, we are going to use a stoichiometric approach from the balanced reaction equation:

5 moles of O₂ will react completely to produce 4 moles of NO

1 mole of O₂ will therefore react to produce x mole of NO

5x = 4

x = $\frac{4}{5}$ moles of NO

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Help! It’s for chemistry I attached a picture thank you!

IgorLugansk [536]

Answer:

0.15 L

Explanation:

You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.

5.00 kg = 5,000 g

(5,000 g)/(1.00 g/mL) = 5,000 mL

5,000 mL = 5 L

Now, find the volume the seawater will take up.

(5,000 g)(1.03 g/mL) = 4854.4 mL

4854.4 mL = 4.85 L

Subtract the two volumes to find the volume that left unfilled.

5 L - 4.85 L = 0.15 L

6 0

2 years ago

Two airplanes are flying in the sky in such a way that the gravitational attractions between them decreases. The planes must be

masya89 [10]

A. away from each other

8 0

3 years ago

The number of proton, neutrons and electrons in oxygen atom

Ksenya-84 [330]

Answer:

in an oxygen atom there are:

protons:8

electrons:8

neutrons:8

Explanation:

this is because the atomic number of oxygen is 8 and that is the proton number and the electron number is the same as the atomic number

4 0

3 years ago

How many molecules are there in 122 grams of Cu(NO3)2?

ipn [44]

Cu =63.5
2 times N =28.02
6 times O =96
96=63.5=28.02=127.07
122/127.07=.96 molecules

6 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago