Answer:
K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not
Explanation:
Find the molar mass of K2SO4 first:
2K + S + 4O ≈ 174 g/mol
Divide the goal molar mass of 696 by the molar mass of the empirical formula:
696 / 174 = 4
This means you need to multiply everything in the empirical formula by 4:
K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not
The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms
<h3>Avogadro's hypothesis </h3>
1 mole of Mg = 6.02×10²³ atoms
<h3>How to determine the atoms in 0.58 mole of Mg </h3>
1 mole of Mg = 6.02×10²³ atoms
Therefore,
0.58 mole of Mg = 0.58 × 6.02×10²³
0.58 mole of Mg = 3.49×10²³ atoms
Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg
Learn more about Avogadro's number:
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Answer:
d= 50.23 g/cm³
Explanation:
Given data:
radius = 137.9 pm
mass is = 5.5 × 10−22 g
density = ?
Solution:
volume of sphere= 4/3π r³
First of all we calculate the volume:
v= 4/3π r3
v= 1.33× 3.14× (137.9)³
v= 1.33 × 3.14 × 2622362.939 pm³
v= 1.095 × 10∧7 pm³
v= 1.095 × 10∧-23 cm³
Formula:
Density:
d=m/v
d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³
d= 5.023 × 10∧+1 g/cm³
d= 50.23 g/cm³
Answer:
22s
Explanation:
Given parameters:
Acceleration = 25m/s²
Final velocity = 550m/s
Initial velocity = 0m/s
Unknown:
Time taken = ?
Solution:
To solve this problem, we are going to use one of the motion equations:
v = u + at
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
550 = 0 + 25 x t
550 = 25t
t = 22s
The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol
The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.
The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2) <span>= -120.535714286 kJ/g or -121 kJ/g
</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g)
<span>= -50 kJ/g
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.
