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2 years ago

What is the last step in the scientific method?

Chemistry

2 answers:

yawa3891 [41]2 years ago

8 0

Answer:

D. Communication of results

Anit [1.1K]2 years ago

4 0

Answer:

conclusion

because it gives the overall information about the experiment

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A compound is 79.94g carbon, 26.82g hydrogen, and 93.24g nitrogen. What is the empirical formula?

Pani-rosa [81]

Answer:

i dont know

Explanation:

sorry

5 0

2 years ago

1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

$\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

$\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0

2 years ago

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

8 0

2 years ago

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

What happens to the particles during solid to a liquid

Ratling [72]

the particles from solid to liquid start to move around faster then it was at the first state

3 0

2 years ago