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lozanna [386]
2 years ago
8

A hypothesis is made before the experiment is conducted. True or fasle

Chemistry
1 answer:
Viefleur [7K]2 years ago
5 0

It should be at beginning. A hypothesis is called an educated guess of what might happen in the experiment.

Please brainliest

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1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
What is the difference between pioneer species and climax communities
bonufazy [111]
Pioneer species are the first species to arrive in an area after succession (hope this helped because i dont know about climax communities)
5 0
3 years ago
20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

To know more about Percentage Yield

brainly.com/question/22257659

#SPJ1

5 0
1 year ago
_____ chemical reactions need a certain amount of activation energy to get started.
Anettt [7]

Answer:all

Explanation:

Because a reaction does not starts by itself unless some force is applied

3 0
3 years ago
Consider the following reaction at 298K.
lakkis [162]

The true statements are;

  • K < 0
  • Eocel  < 0

<h3>What is a redox reaction?</h3>

We define a redox reaction as one in which a specie is oxidized and another is reduced.

Now;

Eo cell = cell potential = -0.13 V - (+0.34 V) = -0.47 V

n =number of moles of electrons = 2 mole of electrons

K = equilibrium constant

ΔG = change in free energy

Eo cell = 0.0592/n log K

-0.47 =  0.0592/2 log K

log K =  -0.47  * 2/0.0592

K = 1.3 * 10^-16

ΔG = -nFEo cell

ΔG = -(2 * 96500 * -0.47)

ΔG = 90.7kJ

Learn more about Ecell:brainly.com/question/10203847

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8 0
2 years ago
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