Answer:
I will need six (6) bags of potting soil
Explanation:
Since you plan on planting in 10 pots and need 15 cups of potting soil per pot, the total amount of potting soil you need <em>(in cups)</em> is 10 X 15 = 150 cups of potting soil.
We have that a bag contains 25 sups. to get the number of bags needed, we have to divide 150 by 25. This will give us 150 / 25 = 6 bags.
Therefore, I will need six (6) bags of potting soil
Please show the whole question including the diagram so that you can receive help with the question.
Explanation:
To find the amount of product that would be formed from two or more reactants, we need to follow the following steps;
- Find the number of moles of the given reactants.
- Then proceed to determine the limiting reactant. The limiting reactant is the one in short supply which determines the extent of the reaction.
- Use the number of moles of the limiting reactant to find the number of moles of the product.
- Then use this number of moles to find the mass of the product
Useful expression:
Mass = number of moles x molar mass
Answer:1.123 x 10^-31cm
Explanation:
mass of humming bird= 11.0g
speed= 1.20x10^2mph
but I mile = 1.6m
1km=1000
I mile = 1.6x10^3m
1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2
=1.932 x10^5m
recall that
1 hr= 60 min
1 min=60 secs, 1hr=3600s
Speed = distance/ time
=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s
m= a 11.0g= 11.0 x 10^-3kg
h=6.626*10^-34 (kg*m^2)/s
Wavelength = h/mu
= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)
6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m
but 1m = 100cm
1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm
de broglie wavelength of humming bird = 1.123 x 10 ^-31cm
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
