4.The other light bulb will stay on and glow brightly.
Answer:- 3333 g of solution.
Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.
(i) 0.045% NaCl by mass
Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?
We could solve this using proportions...
(0.045/100) = (1.5/X)
0.045(X) = 1.5(100)
0.045X = 150
X = 150/0.045 = 3333
So, 1.5 g of NaCl is present in 3333 g of solution.
Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
The average pea weighs between 0.1 and 0.36 grams.
If we take the lower value (0.1 g/pea), the number of peas in 454 g is:
If we take the higher value (0.36 g/pea), the number of peas in 454 g is:
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
You can learn more about conversion factors here: brainly.com/question/1844638
