A conductor that is more powerful and can last longer
Answer:
Explanation:
Entropy is measure of disorder so as we lower the temperature of gas , its entropy decreases .
Hence at - 200°C entropy of nitrogen will be less than that at - 190°C .
At freezing point ,
entropy of fusion = latent heat / freezing temperature
= .71 kJ / ( 273 - 210 )
= 710 / 63 J mol⁻¹ K⁻¹ .
= 11.27 J mol⁻¹ K⁻¹ .
entropy of fusion = 11.27 J mol⁻¹ K⁻¹ .
Answer:
Compounds 1 and 2 are not the same
Explanation:
To solve this question we need to find the molecular formula of the compounds converting the mass of each atom to moles.
<em>Molecular formula is defined as the simplest whole number ratio of atoms present in a molecula:</em>
<em />
Compound 1:
<em>Moles Tin: </em>
5.63g Sn * (1mol / 118.7g) = 0.04743 moles
<em>Moles Cl:</em>
3.37g Cl * (1mol / 35.45g) = 0.09506 moles
Ratio Cl:Sn
0.09506 moles / 0.04743 moles = 2
Molecular formula SnCl₂
Compound 2:
<em>Moles Tin: </em>
2.5g Sn * (1mol / 118.7g) = 0.02106 moles
<em>Moles Cl:</em>
2.98g Cl * (1mol / 35.45g) = 0.08406 moles
Ratio Cl:Sn
0.08406 moles / 0.02106 moles = 4
Molecular formula SnCl₄
Compounds 1 and 2 are not the same because molecular formulas are different.
Answer:
Explanation:
I got you....so equilibrium constant is written as products/reactants with raising the products and reactants to appropriate powers based on leading coefficients.
For eg: A+2B------->2C+3D
Keq for this is products/reactions i.e. [(C)^2 (D)^3]/[(A) (B)^2] you see we raised each product or reactant for leading coefficient (the number in balanced equation) power.
In the question above.....
The keq would thus be therefore answer is B
Answer:
The answer to your question is 40 L of NH₃
Explanation:
Data
Volume of NH₃ = x
mass of N₂ = 25 g
mass of H₂ = excess
Balanced chemical reaction
N₂ + 3H₂ ⇒ 2NH₃
Process
1.- Find the molar mass of N₂ and NH₃
N₂ = 14 x 2 = 28g
2NH₃ = 2[ 14 + 3] = 34 g
2.- Write a proportion to solve this problem
28 g of N₂ --------------- 34 g of NH₃
25 g of N₂ ------------- x
x = (25 x 34)/28
x = 30.36 g of NH₃
3.- Calculate the volume of NH₃
17 g of NH₃ -------------- 22.4 L
30.36 g of NH₃ -------- x
x = (30.36 x 22.4) / 17
x = 40 L