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Tatiana [17]
3 years ago
14

PLEASE HELP ME!!!!!!!!!!

Chemistry
1 answer:
frosja888 [35]3 years ago
6 0

Answer:

hopefully that helps

Explanation:

-24

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The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction
andreev551 [17]

Answer:

The NO + O3 is the dominant reaction.

Explanation:

First of all, let's convert to molecules/cm³;

For O3;

O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT

Thus, plugging in the relevant values to get;

n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)

So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =

9.84 x 10^(11) molecules/cm³

But from the question, NO has 2 moles, and thus concentration is;

2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³

For O2;

Following the same pattern for O3, we obtain;

(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³

Now, for NO and O3 reaction the rate is; k[NO] [O3]

Thus rate;

= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s

For 2NO + O2 → 2NO2 reaction, rate = k[NO]2 [O2]

Thus, rate;

= (2 x 10^(-38) cm^(6)/molec².s )( 1.968 x 10^(12) molec/cm³) ²

(5.167 x 1018 molec/cm³)

= 40,000 molec/cm³.s

Observing the two rates, it's clear that the NO + O3 is the dominant reaction.

6 0
3 years ago
Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar
DIA [1.3K]
Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol


2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g----------------------  x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles--------        x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
4 0
3 years ago
Knowledge of atomic structure has evolved over time. Thomson discovered that atoms contain electrons, Rutherford discovered that
sergejj [24]
I believe the correct response would be A. The cumulative nature of science. How experimental evidence and ideas made by many scientists have accumulated and or built up to have a more understandable and or clear view of a particular scientific topic, and or principle.
4 0
3 years ago
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How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)
marysya [2.9K]

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

 We are to convert from ounces to micrograms.

                                    1ug  = 1 x 10⁻⁶g

                                    1oz  = 28.35g

       

So we first convert to grams from oz then take to ug:

 Solving:

                    1oz  = 28.35g

             3.4 x 10⁻⁵oz  will then give  3.4 x 10⁻⁵ x 28.35 = 9.64  x 10⁻⁴g

So;

                    1 x 10⁻⁶g    = 1ug

          9.64  x 10⁻⁴g will give \frac{9.64 x 10^{-4} }{1 x 10^{-6} }      = 9.64 x 10²ug or 964ug

8 0
2 years ago
As a mixed two substance in a test tube she noticed that the test tube because warm she also noticed gas coming from the test tu
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That is convection my friend
4 0
2 years ago
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