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Sindrei [870]
3 years ago
10

PLEASE HELP! Class is almost ending, I don't want to repeat.

Chemistry
1 answer:
nikklg [1K]3 years ago
6 0
ANSWERS:

1. 2KClO_{3} ⇒ 2KCl + 3O_{2}

2. H_{2}  C_{2} O_{4}  + 2NaOH ⇒ Na_{2}  C_{2} O_{4}  + 2 H_{2} O

3. 2 C_{10}  H_{22} +  31O_{2} ⇒ 20C O_{2} + 22 H_{2} O

4. 4 H_{2} + Fe_{3} O_{4} ⇒ 3Fe +4 H_{2} O

5. 4P + 5 O_{2} ⇒ 2 P_{2}  O_{5}


I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

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tiny-mole [99]
The answer is A, between 0 and 7.

In a pH scale from 0 to 14, we can groups these numbers into acidic, neutral, and alkaline. 7 is the neutral pH value, therefore, 0-7 is always acidic, and 7-14 is alkaline.
The smaller the number is, the more acidic the solution will be. This applies same in alkalis, the larger the pH value is, the more alkaline the solution is.

We can measure the pH of solution with many methods, the easiest way include using a pH paper, more advanced and accurate methods includes using a pH meter.
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melisa1 [442]

Answer:

Farther, longer, longer

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2 years ago
The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

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8 0
3 years ago
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olga55 [171]
I suck at chemistry but i have a friend that can help
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Green light has a wavelength of 550nm. What is the frequency of a photon of green light?
Vikki [24]

Answer:- Frequency is 5.4*10^1^4Hz .

Solution:- frequency and wavelength are inversely proportional to each other and the equation used is:

\nu =\frac{c}{\lambda }

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Speed of light is 3.0*10^8m/s .

We need to convert the wavelength from nm to m.

( 1nm=10^-^9m )

550nm(\frac{10^-^9m}{1nm})

= 5.5*10^-^7m

Now, let's plug in the values in the equation to calculate the frequency:

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= 5.4*10^1^4s^-^1 or 5.4*10^1^4Hz

since, 1s^-^1=1Hz

So, the frequency of the green light photon is 5.4*10^1^4Hz .

4 0
2 years ago
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