Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.
To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:
- 8.89 mol Al(OH)₃ *
= 26.67 mol HCl
Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
I think one of the signs is <span>water and salt are formed
</span>
Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
Answer:
Explanation:
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)
Explanation;
According to Arrhenius equation:
i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)
Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K
T2 = 25 oC = (25 + 273) K = 298 K
i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)
i.e. ln(k2/0.000717) = -2.54738
i.e. k2/0.000717 = 
= 0.078286
Therefore, the required constant (k2) = 0.078286 * 0.000717 = 