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shutvik [7]
3 years ago
8

If 38.7 grams of lithium chlorate decomposes, how many grams of oxygen gas can be produced?

Chemistry
2 answers:
lesantik [10]3 years ago
8 0
The answer to your question is 20.6
solniwko [45]3 years ago
6 0
LiClO3 has mole of mass of 90 g/mol

So 38.7 g means 0.43 mol

Since in this reaction, LiClO3 : O2 = 2:3, then it will produce 0.43*3/2 = 0.645 mol O2

So it is equivalent to 0.645 * 32 = 20.64 g O2
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Someone can help me with the question 3, please
pav-90 [236]
Answer ————
60.8 g ammonia
5 0
3 years ago
Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Fluoride is many orders of m
TEA [102]

Answer:   Bromide is many orders of magnitude better than fluoride in leaving group ability

Explanation:

As Size of an atom  Increases,  the Basicity Decreases this is because  if we move downwards  from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element  will be less likely to act as a base implying that the element will be less likely to share its electrons.

in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group  from

F--- worst

Cl----fair

Br ----good

 I-----excellent

with fluorine having the worst ability to leave than Bromine which is better in terms of the leaving group ability.

3 0
3 years ago
In this reaction, _____.
ella [17]

Answer:

AB + CD ----> AC + BD

Explanation:

If you think this reaction:

AB + CD ----> AC + BD

(Reactants)     (Products)

All the statements are true.

7 0
4 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
How many molecules are in 3.6 grams of NaCl?
raketka [301]

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

6 0
3 years ago
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