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Mice21 [21]
2 years ago
7

How many carbon atoms are there in one molecule of CH3FO?

Chemistry
1 answer:
Vlad1618 [11]2 years ago
8 0

Answer:

Brainliest pls

Explanation:

1 mole of a substance contains Avagadro’s number of particles,

i.e. 6.023*10^23

By unitary method,

5 moles of oxygen contains 5 times the Avagadro’s number of particles

i.e. 5* (6.023*10^23) = 3.0115*10^24 number of particles.

Now, the further answer depends on what particles the question concentrates on.

If number of atoms are asked , the above answer must be multiplied by 2, because oxygen is a diatomic gas and each atom contributes to be a particle.

therefore, 5 moles of oxygen has 6.023*10^24 atoms.

If number of molecules asked, the above answer is directly written...

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Calculate the answer. Express it in scientific notation and include the correct number of significant figures. (12 x 104 ) x (5
ipn [44]

Answer:

Explanation:

(12 x 104 ) x (5 x 10-²) = 6 x 10 ³ 6 x 105 6.0x10²

1.  (12 x 104 ) = 1248 or 1.248 x 10^3

2.  (1.248 x 10^3)(5 x 10^-2) = 6.240 x 10^1 or 60 rounded to 1 sig fig

I don't understand "= 6 x 10 ³ 6 x 105 6.0x10² "

6 0
1 year ago
What is the pressure in atmospheres exerted by a 0.500 mole sample of nitrogen gas in a 10.0 L
Nonamiya [84]

Answer:

The pressure is 1, 22 atm.

Explanation:

We use deal gas formula. First, we convert the unit of temperature in Celsius into Kelvin. We use the constant R= 0,082 l atm /K mol.Then, we solve P (pressure).

0°C=273 K   25°C= 273 + 25= 298 K

PV=nRT   -----> P= (nRT)/V

P= (0,5 mol x 0,082 l atm /K mol x 298 K)/ 10 L

<em>P= 1, 2218 atm</em>

3 0
3 years ago
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
1. How does cellular respiration add carbon to the<br> atmosphere?
asambeis [7]
Glucose and oxygen are changed into energy and carbon dioxide during cellular respiration and that’s how carbon dioxide is released in the air.
8 0
2 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
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