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Ray Of Light [21]
3 years ago
7

Can anyone help me with this?:(

Chemistry
1 answer:
Kobotan [32]3 years ago
3 0

I believe this would be an exothermic reaction, As time increases, the temperature goes up meaning the heat from the system is going into the surroundings. Which indicates an exothermic reaction.

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Determine the molecular
jeyben [28]

Answer:

MgO- magnesium oxide

Cu(NO3)2- copper(11)nitrate

Li2CO3- lithium carbonate

7 0
3 years ago
We have 1700 Tonnes of Amoniac every day. How many tonnes of 63% Nitrogen soda we can get?
V125BC [204]
There are 3 equations involved in manufacturing Nitric Acid from Ammonia. 

First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O

Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4

Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO

Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)

Convert the exact number back into tons, and your answer is: 419.332775 tons.

Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)

Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.
3 0
3 years ago
Which are characteristics of the speed of light? Check all that apply.
valkas [14]
- <span>The </span>speed of light<span> is the </span>speed<span> electromagnetic waves travel. The </span>speed of light<span> is 3.0 x 108 m/s in space. The </span>speed of light<span> is the same in all matter. The </span>speed<span> of higher frequency </span>light <span>slows down more than lower frequency </span>light<span> in matter.</span>
4 0
3 years ago
Read 2 more answers
How many moles of oxygen are required to produce 37.15 g CO2? 37.15 g CO2 = mol O2
NNADVOKAT [17]

Answer:

0.84 moles of oxygen are required.

Explanation:

Given data:

Mass of CO₂ produced = 37.15 g

Number of moles of oxygen = ?

Solution:

Chemical equation:

C + O₂     →     CO₂

Number of moles of  CO₂:

Number of moles = mass/molar mass

Number of moles = 37.15 g/ 44 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of oxygen and carbon dioxide.

                          CO₂         :       O₂  

                              1           :         1

                            0.84       :       0.84

0.84 moles of oxygen are required.

6 0
3 years ago
Read 2 more answers
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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