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mojhsa [17]
3 years ago
8

Which of the following is always a reactant in a combustion reaction?

Chemistry
2 answers:
Margaret [11]3 years ago
6 0
I believe it’s oxygen

Marina86 [1]3 years ago
5 0

Answer:

Oxygen

Explanation:

Any Hydrocarbon + Oxygen ----> Carbon Dioxide + H2O (Water vapour, gas, water)

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Phosphorus can be stable with 12 electrons in its valence structure while nitrogen can never have more than 8 electrons in its v
dolphi86 [110]

Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.

Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.

On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.

8 0
2 years ago
Please help :( I don’t understand this question
Karolina [17]

Answer:

A...................................

8 0
3 years ago
Read 2 more answers
Convert 11.03 moles of calcium nitrate to grams
hichkok12 [17]
Im pretty sure the answer is <span> 0.01218859659g

not 100% sure tho so please consult someone else b4 answering 
i hope this helps!!</span>
6 0
3 years ago
Read 2 more answers
What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

3 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
2 years ago
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