Answer:
the pressure exerted in pascals is 0.5 Pa
Explanation:
The computation of the pressure exerted in pascals is shown below:
As we know that
Pressure = force ÷ area
= 25 ÷ 50
= 0.5 Pa
Hence, the pressure exerted in pascals is 0.5 Pa
We simply applied the above formula so that the correct pressure could come
Answer:
The molar mass of a compound is The mass in grams of 1 mole of the compound (Option A)
Explanation:
Let's take ammonia as an example (NH3)
Mass of N = 14 g
Mass of H = 1 g
Molar mass of ammonia is Mass of N + (Mass of H).3
14 + 3 = 17 g/m
Ammonia is a compound that has 1 mol of N, plus 3 moles of H (see the formula)
The number of atoms in 1 mole of the compound --> This is Avogadro
Answer:
Explanation:
Metal elements form positively charged ions called cations because they are located on the left side of the periodic table These elements all have valence electrons in an s orbital. These electrons are relatively easy for the atom to lose to achieve a stable octet of electrons in its outermost energy shell.
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10
