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3 years ago

Get it right :D if this is wrong or not correct ill give u a bad review And no brainiest

Chemistry

1 answer:

tangare [24]3 years ago

4 0

Answer:

In short, because we are tearing up the oxygen factories to make way for carbon dioxide emitters. (Doesn't make a lot of sense, read the explanation)

Explanation:

So, 1000 years ago, we had a lot more trees, didn't have engines or cars or factories or anything, really that released carbon dioxide into the air and we had a lot more trees and since the invention of cars, engines, carbon dioxide-emitting tools and factories and all the other things that emit "Greenhouse gases" and in doing that, cleared more trees to make room for factories and roads and that has drastically changed the outlook of the carbon cycle.

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Explain how the crystals of an igneous rock are used to help classify it?

Hatshy [7]

Classify what. explain more

6 0

3 years ago

1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.

Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

$H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ$ ..[1]

$CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ$ ..[2]

$C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ$ ..[3]

$2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?$ ..[4]

2 × [2] - [1]- [3] = [4] (Using Hess's law)

$\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3$

$\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)$

$\Delta H_4=64.9 kJ/mol$

The enthalpy of the reaction is 64.9 kJ/mol.

3 0

3 years ago

How many liters of oxygen gas, at standard

Karo-lina-s [1.5K]

Answer:

Explanation:

For the balanced reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

Using cross multiplication:

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

Finally, we can get the volume of oxygen using the information:

It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).

Using cross multiplication:

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0

3 years ago

Radio waves travel at the speed of light which is 3.00 x 10^8. How many minutes does it take for a radio message to reach saturn

ryzh [129]

Answer:

43.89 min

Explanation:

Given that:-

The speed of light = $3.00\times 10^8\ m/s$

The distance = $7.9\times 10^8\ km$

The conversion of distance in km to distance into m is shown below as:-

1 km = 1000 m

So,

Distance = $7.9\times 10^8\times 1000\ m=7.9\times 10^{11}\ m$

The relation between speed distance and time is shown below as:-

$Speed=\frac{Distance}{Time}$

Thus,

$3.00\times 10^8=\frac{7.9\times 10^{11}}{Time}$

$300000000\times time=10^{11}\times \:7.9\ s$

Time = 2633.33 seconds

Also, 1 s = 1/60 min

So,

Time= $\frac{2633.33}{60}\ min=43.89\ min$

3 0

3 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

8 0

3 years ago