Answer is: mass of sodium hydroxide is 60 grams.
V(NaOH) = 300 mL · 1 L/1000mL = 0,3 L.
c(NaOH) = 5 M = 5 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 5 mol/L · 0,3 L.
n(NaOH) = 1,5 mol.
m(NaOH) = n(NaOH) · M(NaOH).
m(NaOH) = 1,5 mol · 40 g/mol.
m(NaOH) = 60 g.
The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.
Explanation:
Answer is "C".
<em><u>Explanation
</u></em>
Single replacement reaction is a type of reaction which one reactant reacts with another and makes a product by replacing one element by another.
Mg (Magnesium) reacts with Al₂O₃ (Aluminium oxide) and produces MgO (Magnesium oxide) and Al (Aluminium) as products. Here Al is replaced by Mg. Reaction is
Mg + Al₂O₃ → MgO + Al
To balance the reaction equation, both left and right hand sides should have same number of atoms in each element.
Here,
<em>Left Hand Side has </em> <em>Right Hand Side has</em>
Mg = 1 atom Mg = 1 atom
Al = 2 atom Al = 1 atom
O = 3 atom O = 1 atom
First step : balance the O atoms in both sides. To do that "3" should be added before MgO.
Second step : After balancing O atoms, there will be 3 Mg atoms in right hand side. Hence to balance Mg atoms again "3" should be added before Mg in left hand side
Third step : as the final step balance the Al atoms by adding "2" before Al in the right hand side.
Then final balanced equation should be
3Mg + Al₂O₃ → 3MgO + 2Al
<span>A machine is a device that does work. Most machines consist of a number of elements, such as gears and ball bearings, that work together in a complex way. Nonetheless, no matter how complex they are, all machines are based in some way on six types of simple machines. These six types of machines are the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and t</span>
Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
