Answer:
the answer is D
Explanation:
Air over the beach heats up, rises, and is replaced by cool, denser ocean air
The ocean heats up much slower and becomes relatively cooler than the beach making it a high pressure zone. The air moves from the water to the land each morning forming a sea breeze. These pressure zones might equal out later in the day and the winds may diminish. ... A sea breeze blows from the ocean towards the beach.
Answer:
Weight of element X = 14.01 gram
Explanation:
Mass percentage is calculated by using :


Mass of O = 16.0 gram
let mass of X = X grams = ?
Mass of XO2 = mass of X + 2(mass of O)
Mass of XO2 = X + 2(16)
Mass of XO2 = X + 32
Total mass of O in the compound = 2(16) = 32 grams
Percent O = 69.55 %









weight of element X = 14.01 gram
14.01 is mass of nitrogen N
The correct formula of this element is NO2
Answer:
Hydrosulfuric acid will act as limiting reactant.
Explanation:
Given data:
Mass of iron(III) chloride = 3243.0 g
Mass of hydrosulfuric acid = 511.8 g
Limiting reactant = ?
Solution:
Chemical equation:
2FeCl₃ + 3H₂S → Fe₂S₃ + 6HCl
Number of moles of iron(III) chloride:
Number of moles = mass/molar mass
Number of moles = 3243.0 g/ 162.2 g/mol
Number of moles = 20 mol
Number of moles of hydrosulfuric acid:
Number of moles = mass/molar mass
Number of moles = 511.8 g/ 34.1 g/mol
Number of moles = 15 mol
Now we will compare the moles of both reactant with products
FeCl₃ : Fe₂S₃
2 : 1
20 : 1/2 ×20 = 10
FeCl₃ : HCl
2 : 6
20 : 6/2 ×20 = 60
H₂S : Fe₂S₃
3 : 1
15 : 1/3 ×15 = 5
H₂S : HCl
3 : 6
15 : 6/3 ×15 = 30
Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.
Answer:
half lives passed=5
given sample=90g
sample left=2.8125g
Explanation:
no. of half lives=total time/half life
no.=19days/3.8days
no.=5 days
after 5 half lives sample left=2.8125g
Answer:
58.44 g of NaCl are needed.
Explanation:
Given data:
Mass of NaCl needed = ?
Volume of solution = 200 mL (200/1000 =0.2 L)
Molarity of solution = 5 M
Solution:
We will solve this problem through molarity formula.
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will put the values.
5 M = moles of solute / 0.2 L
Moles of solute = 5 mol/L × 0.2 L
Moles of solute = 1 mol
Mass of sodium chloride:
Mass = number of moles × molar mass
Mass = 1 mol × 58.44 g/mol
Mass = 58.44 g
Thus, 58.44 g of NaCl needed.