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MrMuchimi
3 years ago
14

Describe how hydrogen and oxygen form water

Chemistry
2 answers:
Rama09 [41]3 years ago
8 0
Just google it and put it into your own words
Gwar [14]3 years ago
7 0

Answer:

The  hydrogen molecules combine with the oxygen molecules, 2 hydrogen molecules, and 1 oxygen molecules is the amount needed to make one water atom or molecule, whatever you want to call it.

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We can consider a chemical to be safe if:
slamgirl [31]
We can consider a chemical to be safe if it does not contain any harmful substances such as nitrogen, harmful acids, or even excessive heat. Other chemicals such as the chemicals used in food products like, for example, citric acid, is not so harmful.
Hope this helps! :D
Any questions? Just let me know! I'd be happy to help any way possible.
8 0
3 years ago
Which molecule in plants cells first captures the radiant energy from sunlight?
melisa1 [442]

Answer:

c tell  me if im wrong:)

Explanation:

:)

7 0
2 years ago
If 12.4 mol of Ne gas occupies 122.8 L, how many mol of Ne would occupy 339.2 L under the same temperature and pressure? Record
8090 [49]

Answer:

3.43×10¹ mol

Explanation:

Given data:

Initial number of  moles = 12.4 mol

Initial volume = 122.8 L

Final number of moles = ?

Final volume = 339.2 L

Solution:

The number of moles and volume are directly proportional to each other at same temperature and pressure.

V₁/n₁  =  V₂/n₂

122.8 L/ 12.4 mol  =  339.2 L / n₂

n₂ = 339.2 L× 12.4 mol  / 122.8 L

n₂ = 4206.08 L.mol /122.8 L

n₂ = 34.3mol

In scientific notation:

3.43×10¹ mol

7 0
3 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
3 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
Read 2 more answers
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