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IgorC [24]
3 years ago
14

Nitrous oxide (N2O) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is also a potent gr

eenhouse gas. The decomposition of N2O is first order. It decomposes slowly to N2 and O2:
2 N2O(g) --> 2 N2(g) + O2(g)

How many half-lives will it take for the concentration of the N20 to reach 3.13% of its original concentration?
Chemistry
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:

five half lives

Explanation:

Half-life is the time required for a quantity to reduce to half of its initial value.

How many half lives it would take to reach 3.13% form 100% of it's initial concentration:

100% - 50% : First Half life

50% - 25%: Second Half life

25% - 12.5%: Third Half life

12.5% - 6.25%: Fourth Half life

6.25% - 3.125%: Fifth Half life

This means it would take five half lives to get to 3.125% (≈ 3.13%) of it's original concentration.

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Use the formula for pressure to determine the weight of a box that creates a pressure of 1,065 N/cm 2 when resting on a 17.0-cm
S_A_V [24]
Force /Area = Pressure
Force = Pressure * Area
<span>Force = 1065 * 17 = 18105 N</span>
7 0
3 years ago
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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
3 years ago
Is the law of conservative mass observed in this equation CaCO3 + 2HCI --&gt;CaCI2 +H2O + CO2
pychu [463]

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:

  • Ca ⇒ 1 on the left, 1 on the right.
  • C ⇒ 1 on the left, 1 on the right.
  • O ⇒ 3 on the left, 3 on the right.
  • H ⇒ 2 on the left, 2 on the right.
  • Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

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Use the scenario to answer the following question. Four scientists observed an area of land that had been cleared of all trees a
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Answer:

b

Explanation:t not sure

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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
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Answer is: glycerol because it is more viscous and has a larger molar mass.

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The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
3 years ago
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