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Elan Coil [88]
3 years ago
11

16 mol of oxygen were react with excess hydrogen gas, how many moles of water would be produced

Chemistry
2 answers:
nydimaria [60]3 years ago
8 0

Answer: 32 moles of water.

Explanation:

The balanced chemical equation for formation of water:

2H_2+O_2\rightarrow 2H_2O

As can be seen from the chemical equation, 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

As it is given that hydrogen is present in excess, hydrogen is the excess reagent and oxygen is the limiting reagent as it limits the formation of product.

According to stoichiometry:

1 mole of Oxygen gives 2 moles of water

Thus 16 moles of oxygen will give=\frac{2}{1}\times 16=32molesof water.

32 moles of water will be produced.

german3 years ago
7 0
2H₂  +  O₂   →   2H₂O

mole ratio of O₂  :  H₂O     is     1  :  2

∴  if moles of O₂ = 16 mol

then moles of H₂O = (16 mol ×  2)

                               =  32 mol
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A 31.5 g wafer of pure gold initially at 69.4 ∘C is submerged into 63.4 g of water at 27.4 ∘C in an insulated container.
liubo4ka [24]

Answer: The final temperature of both substances at thermal equilibrium is 301.0 K

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of gold = 31.5 g

m_2 = mass of water = 63.4 g

T_{final} = final temperature = ?

T_1 = temperature of gold = 69.4^oC=342.4K

T_2 = temperature of water = 27.4^oC=300.4K

c_1 = specific heat of gold = 0.129J/g^0C

c_2 = specific heat of water= 4.184J/g^0C

Now put all the given values in equation (1), we get

-31.5\times 0.129\times (T_{final}-342.4)=[63.4\times 4.184\times (T_{final}-300.4)]

T_{final}=301.0K

The final temperature of both substances at thermal equilibrium is 301.0 K

4 0
4 years ago
Complete the nuclear reaction.*
irina [24]

\boxed {_{-1}^0e}

Explanation:

\huge _{53}^{125}I\rightarrow _{54}^{125}Xe+\boxed {_{-1}^0e}

5 0
3 years ago
32 Points!!! Match the definition with correct keyword.
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Answer: Okay so here's the order lol from top to bottom

2, 1, 3, 4, 5

Explanation:

5 0
3 years ago
When writing an electron configuration for a transition metal, what is the last orbital type to be filled – an s, p, d or f orbi
Black_prince [1.1K]
It would be d.

Reason being said...

the electron configuration normally goes like this...

1s2 2s2 2p6 3s2 3p6 4s2....

until you hit the transition metals..remember those have a special rule..

even though you are in the 4 sublevels for the orbitals ... it goes down 1

Making it 3d..(1,2,3,4,5,6,7,8,9,10)

Going on...

at 5s2 then, 4d1, 4d2, 4d3, 4d4, etc..

at 6s2 then, 5d1, 6d2, 6d3, 6d4, etc..

Thus, D orbital is your answer.
4 0
4 years ago
Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate
Anvisha [2.4K]

Answer:

-3272     kJ/mol

Explanation:

Given and known facts

Mass of Benzene = 0.187 grams

Mass of water = 250 grams

Standard heat capacity of water = 4.18 J/g∙°C

Change in temperature ΔT = 7.48°C

Heat

=250 * 4.18 * 7.48\\=7816.6 \\=7.82

Heat released by benzine is - 7.82 kJ

Now, we know that

0.187 grams of benzene release = -7.82  kJ heat

So, 1 g benzine releases

\frac{ -7.82 }{0.187}\\= -41.8

kJ/g

0.187 * \frac{1}{78.108}=0.00239 mol C6H6

Heat released

= \frac{-7.82}{ 0.00239}

=-3272     kJ/mol

4 0
3 years ago
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