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3 years ago

10

At a football game, the crowd begins a stadium wave, rising from their seats and falling back to their seats, causing the wave t

o travel around the stadium.

1 answer:

jarptica [38.1K]3 years ago

3 0

What is the problem in the question??

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If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr

Fynjy0 [20]

Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.

7 0

2 years ago

You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physics. Y

lbvjy [14]

Answer: 2.58 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = age of sample = 6 days

a = initial amount of the reactant = 1 g

a - x = amount left after decay process = 0.2 g

a) to find the rate constant

$6=\frac{2.303}{k}\log\frac{1.0}{0.2}$

$k=\frac{2.303}{6}\log\frac{1.0}{0.2}$

$k=0.268days^{-1}$

b) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.268}=2.58days$

The half life is 2.58 days

3 0

2 years ago

Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h

zubka84 [21]

1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3

5 0

3 years ago

Can u answer all of em

TiliK225 [7]

Answer:

one is across

Explanation:

8 0

2 years ago

If hydrofluoric acid is a stronger acid than acetic acid, which statement is most likely true?

Alchen [17]

Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.

Explanation:

A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.

Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.

For example, $HF \rightleftharpoons H^{+} + F^{-}$

The conjugate base is $F^{-}$ which is a weak base.

Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.

$CH_{3}COOH \rightarrow CH_{3}COO^{-} + H^{+}$

Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.

4 0

2 years ago