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nexus9112 [7]
2 years ago
13

For the following example, list the given and unknown information (including gratis or moles)

Chemistry
1 answer:
Setler79 [48]2 years ago
6 0

Answer:

9.6 moles O2

Explanation:

I'll assume it is 345 grams, not gratis, of water.  Hydrogen's molar mass is 1.01, not 101.

The molar mass of water is 18.0 grams/mole.

Therefore:  (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).

The balanced equation states that:  2H20 ⇒ 2H2 +02

It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.

get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2

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H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ
Arisa [49]

Answer: help please !!

Explanation:

4 0
3 years ago
Read 2 more answers
Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa
sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

k = A e^\frac{-E_{a}}{RT}

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>

<em></em>

I hope it helps!

8 0
3 years ago
My teacher is grading this soon can someone help me ASAP!
Stolb23 [73]

10. You demonstrated the difference in density of the two objects. It is a physical property.

11. First calculate the density for all of them: density = mass/volume

Density:

A. 5/6 g/ml

B. 10/9 g/ml

C. 15/16 g/ml

D. 20/10 g/ml

If the density of the substance is higher than the density of the substance it is put in, then it will sink. So substances B and D will sink in water, as their densities are higher than 1 g/ml.

12. Ammonia weighs less than water does-- for example, the weight of 8 gallons of ammonia will be equivalent to the weight of 5 gallons of water.

Hope this helped!

3 0
3 years ago
Which chemical formula corresponds to this structural formula?
lys-0071 [83]
I think the answer is choice D
8 0
3 years ago
Read 2 more answers
A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod
sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

7 0
1 year ago
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