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pashok25 [27]
3 years ago
9

I need this answer. any thing will help

Chemistry
1 answer:
Licemer1 [7]3 years ago
8 0

Answer: condensaytion

Explanation: its reverse vaporation instead of water turning into gas gas turns into water

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How many moles of salt are in 13.8 g of sodium chloride?
Hoochie [10]

Answer: 0.24 moles

Explanation:

Molecular Mass of NaCl (23 + 35.5) = 58.5g

58.5g of Sodium Chloride -------> 1 mole of NaCl

∴ 13.8g of Sodium Chloride  ------>  1 ÷58.5 x  13.8 = 0.2358974  ≈    0.24moles

                                                         -  

3 0
3 years ago
Read 2 more answers
Just wanted to make sure I have the right idea.
Rasek [7]

Explanation:

yes you got the right idea

4 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
Read 2 more answers
What is the daughter nuclide when 0-15 experiences positron emission?
Helen [10]

The daughter isotope (a decay product)of O-15 = N-15(Nitrogen 15)

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

  • alpha α particles ₂He⁴
  • beta β ₋₁e⁰ particles
  • gamma particles γ
  • positron particles ₁e⁰

O-15 emits positron particles ₁e⁰, so the atomic number decreases by 1, the mass number is the same

Reaction

 \tt _8^{15}O\Rightarrow _7^{15}N+_1^0e

The mass number of the daughter isotope = 15, atomic number = 7

If we look at the periodic system, the element with atomic number 7 is Nitrogen (N)

5 0
3 years ago
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
3 years ago
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