Answer:
Explanation:
Magnetic materials are always made of metal, but not all metals are magnetic. Iron is magnetic, so any metal with iron in it will be attracted to a magnet. Steel contains iron, so a steel paperclip will be attracted to a magnet too. Most other metals, for example aluminium, copper and gold, are NOT magnetic.
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;

If
&





E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
Hey there!
Find molar mass of C₆H₂Cl₄.
C: 6 x 12.01
H: 2 x 1.008
Cl: 4 x 35.45
------------------
215.876g
This is the mass of 1 mole of C₆H₂Cl₄.
Multiply 215.876 by 5.99.
215.876 x 5.99 = 1293
The mass of 5.99 moles of C₆H₂Cl₄ is 1293 grams.
Hope this helps!
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Answer:
Molecular formula of cyanogen is C₂N₂
Explanation:
We apply the ideal gases law to find out the mole of cyanogen
P . V = n. R. T
Firstly let's convert the pressure in atm, for R
750 mmHg = 0.986 atm
25°C + 273 = 298K
0.986 atm . 0.714L = n . 0.082 L.atm/mol.K .298K
(0.986 atm . 0.714L) / (0.082 L.atm/mol.K .298K) = n
0.0288 mol = n
Molar mass of cyanogen = mass / mol
1.50 g /0.0288 mol = 52.02 g/m
Let's apply the percent, to know the quantity of atoms
100 g of compound contain 46.2 g of C and 53.8 g of N
52.02 g of compound contain:
(52.02 . 46.2) / 100 = 24 g → 2 atoms of C
(52.02 . 53.8) / 100 = .28 g → 2 atoms of N