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Gnoma [55]
2 years ago
5

NEED HELP ASAP! Write a statement explaining your claim about whether or not the equation is correctly balanced.

Chemistry
2 answers:
cupoosta [38]2 years ago
8 0

Answer:

I think that it it correctly balanced that is my opinion and, because  the way it is set up, that the answer will tell you weather or not it is correctly balanced or not.

Kisachek [45]2 years ago
4 0

Evaluating the number of atoms of each element on the product and reactant side, we can conclude that the student's equation isn't balanced.

<u>Given the chemical reaction</u> :

Al_{4}C_{3} + 6H_{2}O ---> 3CH_{4} + 4Al(OH)_{3}

For a balanced equation, the number of atoms of each element in the product side must be equal to that on the reactant side.

<u>Number of hydrogen</u> :

Reactant side = (6 × 2) = 12

Product side = (6 × 2) + (6 × 2) = 12 + 24 = 24

<em>Product side ≠ reactant side</em> ; Hence, the equation isn't balanced

<u>The balanced equation for the reaction goes thus</u> :

Al_{4}C_{3} + 12H_{2}O ---> 3CH_{4} + 4Al(OH)_{3}

<u>Reactant side</u> :

Al = 4

H = (12 × 2) = 24

O = 12

C = 3

<u>Product side</u> :

H = (6 × 2) + (6 × 2) = 12 + 24 = 24

Al = 4

O = 4 × 3 = 12

C = 3

Therefore, the student's equation isn't balanced, the balanced version of the equation has been given .

Learn more :brainly.com/question/2720947

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the valence electrons of an atom of which element would feel a smaller effective nuclear charge than the valence electrons of a
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D. Strontium is the right answer
5 0
3 years ago
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?
SIZIF [17.4K]
The chemical reaction would be written as follows:

2Al + 3Cl2 = 2AlCl3

We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:

19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
5 0
3 years ago
Read 2 more answers
At 0 degrees Celsius, a gas occupies 22.4L. How hot must the gas be in celcius to reach a volume of 25.0L
NikAS [45]

Answer:

31.7 °C

Explanation:

Charles law states that for volume of a gas is directly proportional to the absolute temperature for a fixed amount of gas at constant pressure

we can use the following equation

V1/T1 = V2/T2

where V1 is volume and T1 is temperature at first instance

V2 is volume and T2 is temperature at second instance

temperature should be in kelvin scale

T1 - 0 °C + 273 = 273 K

substituting the values in the equation

22.4 L / 273 K = 25.0 L / T2

T2 = 304.7 K

temperature in celcius is - 304.7 K - 273 = 31.7 °C

the gas must be 31.7 °C to reach a volume of 25.0 L

7 0
3 years ago
Which of the following elements is a transition metal?
GenaCL600 [577]

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

5 0
3 years ago
Read 2 more answers
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
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