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3 years ago

In what important way was Mendeleev’s Periodic Table different from the modern Periodic Table?

Chemistry

1 answer:

horrorfan [7]3 years ago

6 0

He ordered elements based on their atomic mass instead of their number of protons

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What is the difference in energy between an electron in 1s and an electron in 4s.

Andre45 [30]

Answer:

Explanation:

An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.

5 0

3 years ago

when carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4g of carbon were burned in the presence of

SpyIntel [72]

Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2

7 0

3 years ago

What are isotopes please help

NikAS [45]

Answer:

Isotopes are atoms of the same element with the same number of protons but different number of neutrons.

Hope this helps.

6 0

3 years ago

Read 2 more answers

Calculate the percent composition of the following:<br><br> A. VO3<br><br> B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago

What volume of propane (C3H8) is required to produce 165 liters of water according to the following reaction? (All gases are at

inna [77]

Answer:

We need 41.2 L of propane

Explanation:

Step 1: Data given

volume of H2O = 165 L

Step 2: The balanced equation

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Step 3: Calculate moles of H2O

1 mol = 22.4 L

165 L = 7.37 moles

Step 4: Calculate moles of propane

For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane

Step 5: Calculate volume of propane

1 mol = 22.4 L

1.84 moles = 41.2 L

We need 41.2 L of propane

6 0

3 years ago