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3 years ago

What did the periodic table help scientists discover?

Chemistry

2 answers:

noname [10]3 years ago

6 0

Answer:

Would be letter C

Elements not yet known

Natalija [7]3 years ago

4 0

Answer:

A_________________________

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Explain the three laws of chemical reaction in detain gilling examples

kotykmax [81]

Answer:

The three laws of Chemical Reaction are .The law of constant proportions. The law of multiple proportions. The law of reciprocal proportions.

A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by mass.

potassium and chlorine gas ---> chloride.

Hope this helps, have a good day.✌

8 0

2 years ago

What is safer to drink 113 ml of HCl or 244 ml of NaOH

skad [1K]

Answer:

bleach

Explanation:

idrk

5 0

3 years ago

A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

V₂ = 4,00L

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

V₂ = 16,00L

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

V₂ = 12,31L

I hope it helps!

6 0

3 years ago

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) You find the reaction to be first

Vlada [557]

Answer:

Explanation:

FIND THE SOLUTION IN THE ATTACHMENT

4 0

3 years ago

Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3

Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0

3 years ago