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kompoz [17]
3 years ago
5

How many CO2 molecules in 0.0189 mol of CO2?

Chemistry
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

option C is correct = 1.14 × 10²² molecules of  CO₂

Explanation:

Given data:

Number of moles of CO₂ = 0.0189 mol

Number of molecules = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

For given question:

1 mole of CO₂ =  6.022 × 10²³ molecules of  CO₂

0.0189 mol of CO₂ ×  6.022 × 10²³ molecules of  CO₂ / 1mol

1.14 × 10²² molecules of  CO₂

Thus, option C is correct.

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4 years ago
(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

3 0
3 years ago
If a container contains 3.5 moles and has a volume of 1.5 L, then what volume is it when it contains 8 moles? *
Lilit [14]
Volume would be 3.4 L
8 0
3 years ago
Write balanced equations for the following reactions
Arte-miy333 [17]

Answer:

See explanation

Explanation:

The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;

238/92U + 4/2 He -------> 241/94Pu + 1/0 n

238/92U + 4/2 He ------> 241/94Pu + 1/0 n

14/7N + 4/2 He------> 17/8O + 1/1 p

56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He

4 0
3 years ago
Which of these is an isoelectronic series? question 7 options: 1) na+, k+, rb+, cs+ 2) k+, ca2+, ar, s2– 3) na+, mg2+, s2–, cl–
-BARSIC- [3]
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:

Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6, Ar = 18

A. Na⁺: 11-1 = 10 electrons
     K⁺: 19 - 1 = 18 electrons
     Rb⁺: 37-1 = 36 electrons

B. K⁺: 19 - 1 = 18 electrons
    Ca²⁺: 20 - 2 = 18 electrons
    Ar:  18 electrons
    S²⁻:  16 +2 = 18 electrons

C. Na⁺: 11-1 = 10 electrons
    Mg²⁺: 12 - 2 = 10 electrons
     S²⁻:  16 +2 = 18 electrons

D. Li=3 electrons
    Be=4 electrons
    B=5 electrons
    C=6 electrons

The answer is letter B.
6 0
3 years ago
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