[He] 2s² 2p is the electron configuration of Select one: a. Ne b. P с. О d.s

I have no idea how to do this and need help

den301095 [7]

Answer:

60.55%

Explanation:

nO3=155/48

nO2 used: 3/2nO3=4.84375

percent yield: 4.84375/8=60.55%

5 0

2 years ago

Why do we have to pay if its just learning

Oxana [17]

Answer:

what are we paying for?

Explanation:

8 0

2 years ago

For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac

barxatty [35]

First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3:

when 245.2 KClO3 gives → 96 g O2

2.72 g KClO3 gives → X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

= 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

= 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

83.6 Kg KClO3 gives → X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

= 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

= 8.8 mg of O2

7 0

3 years ago

What is the mass of one mole of carbon-12?

velikii [3]

Answer:

Hello

12 grams

The mass of one mole of carbon-12 atoms is 12 grams.

Hope it helps You.....

Explanation:

4 0

2 years ago

Read 2 more answers

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

2 years ago