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nataly862011 [7]
4 years ago
8

How many atoms is in h20

Chemistry
2 answers:
Nataly_w [17]4 years ago
5 0

Answer:

3 atoms

Explanation:

2 hydrogen atoms and 1 oxygen atom

kow [346]4 years ago
5 0

Answer:3

Explanation:

There are 2 hydrogen atoms and 1 oxygen atoms which makes 3 total

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MrRissso [65]
The Krebs cycle uses acetyl CoA as a reactant.
4 0
3 years ago
Greg has 36 DVDs in his collection.
sp2606 [1]

Answer:

14

Explanation:

4 0
4 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
How is 0.00036 written in scientific notation?
natta225 [31]
3.6 x 10^-4
I am writing this sentence because the answer does not reach the 20 character limit. Thank you have nice day! :D Cx :P
4 0
3 years ago
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