The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
1 is b 2 is a 3 is d 4 is a 5 is c
<u>Answer:</u>
<u>For a:</u> The balanced equation is 
<u>For c:</u> The balanced equation is 
<u>Explanation:</u>
A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.
The given unbalanced equation follows:

To balance the equation, we must balance the atoms by adding 2 infront of both
and
and 3 in front of 
For the balanced chemical equation:

The given balanced equation follows:

The given equation is already balanced.
The given unbalanced equation follows:

To balance the equation, we must balance the atoms by adding 2 infront of 
For the balanced chemical equation:
The given balanced equation follows:

The given equation is already balanced.
91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Explanation:
2NaN3======2Na+3N2
This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.
From the equation:
2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.
In the question moles of nitrogen produced is given as 2.104 moles
so,
From the stoichiometry,
3N2/2NaN3=2.104/x
= 3/2=2.104/x
3x= 2*2.104
= 1.4 moles
So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.
From the formula
no of moles=mass/atomic mass
mass=no of moles*atomic mass
1.4*65
= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Alkaline Earth Metals are the elements located in the second period from the left of the periodic table. These elements lose two electrons to form the stable octet when forming an ionic bond, resulting in a net charge of +2. Because they’re trying to get rid of those electrons to get to the stable octet, it’s easy to remove them - this means that the ionization energy of these elements is relatively low. Finally, since they’re looking to get rid of electrons, they certainly aren’t trying to gain any, meaning that their electronegativity is relatively low.
The correct answers are A and D.