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2 years ago

2Na + 2H2O → 2NaOH + H2

Chemistry

2 answers:

Zarrin [17]2 years ago

5 0

The initial quantity of sodium metal is 17.25

Alinara [238K]2 years ago

5 0

Answer: 13 grams

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to stoichiometry,

1 mole of $H_2$ is produced from 2 moles of sodium

or 22.4 L of $H_2$ at STP is produced from =2 moles of sodium

thus 6.30 L of $H_2$ at STP is produced from = $\frac{2}{22.4}\times 6.30=0.56moles$ of sodium

To calculate the mass of sodium metal used

$\text{mass of sodium used}=moles\times {\text {Molar mass}}=0.56\times 23=13grams$

Thus the initial quantity of sodium metal used is 13 grams.

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Aqueous aluminum bromide and solid zinc are produced by the reaction of solid aluminum and aqueous zinc bromide . write a balanc

Rashid [163]

3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) The unbalanced equation is: ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s) First, count the atoms of each element on each side of the equation: Zn 1,1 Br 2,3 Al 1,1 The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting: 3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s) Now check the atom counts again for both sides: Zn 3,1 Br 6,6 Al 1,2 Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s) Now check the atom counts again: Zn 3,3 Br 6,6 Al 2,2 And they match. So the balanced equation is: 3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)

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2 years ago

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

schepotkina [342]

Answer:

The right solution is " $8.5\times 10^{-19} \ joule$ ".

Explanation:

As we know,

1 mole electron = $6.023\times 10^{23} \ no. \ of \ electrons$

Total energy = $513 \ KJ$

= $513\times 1000 \ joule$

For single electron,

The amount of energy will be:

= $\frac{513\times 1000}{(6.023\times 10^{23})}$

= $8.5\times 10^{-19} \ joule$

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2 years ago

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y (1)

Where X is molar fraction of Al and Y is molar fraction of Li.

X + Y = 1 (2)

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= 3,47%

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1 year ago

As ice transforms into steam, A)new molecules are formed. B)the types of atoms are changed. C)energy is absorbed by the molecule

ra1l [238]

This is an example of a substance changing state, a physical change, the molecules are changed, but the atoms themselves do not change, just their arrangement, and the mass of the molecules is the same. Therefor energy is absorbed by the molecules, as energy is required to change the state or physicality of a molecule structure.Hope this helps. Let me know if you need additional help!

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2 years ago

Read 2 more answers

. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst

Vesnalui [34]

2 KClO3(s) → 3 O2(g) + 2 KCl(s)

Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.

molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028

Moles of O2 produce = $\frac{3}{2} \times 0.028$

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042 = 1.35 g

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2 years ago