Ill say <span>calcium hydroxide but hope this helps :)</span>
Answer:
The sun radiates energy in all directions.
Explanation:
Most of it dissipates into space, but the tiny fraction of the sun's energy that reaches Earth is enough to heat the planet and drive the global weather system by warming the atmosphere and oceans
Answer:
The answer is "6.52 kg and 13.1 kg"
Explanation:
For point a:
Equation:
Calculating the amount of

For point b:

Equation:
Calculating the amount of

Answer:
Density of the object is 8759.494 grams/L
Explanation:
As we know density of an object is mass of the object divided by its volume
Given
Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.
Hence, volume of object is equal to
mL
Mass of the object is
grams
Density
grams/L
Density of the object is 8759.494 grams/L
Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL