Answer:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjkwv-cqrjnAhVCheAKHWaFBBgQFjAAegQICBAB&url=https%3A%2F%2Fwww.acs.org%2Fcontent%2Fdam%2Facsorg%2Feducation%2Fresources%2Fk-8%2Finquiryinaction%2Fstudent-activity-sheets%2Fgrade-5%2Fchapter-3%2Flesson-3.3-forming-a-precipitate.pdf&usg=AOvVaw1fT7fpXG9PNWroM87puvgQ
Explanation:
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Answer:
Specific heat of alloy = 0.2 j/ g.°C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Given data:
Mass of bold = 25 g
Heat absorbed = 250 J
Initial Temperature = 25°C
Final temperature = 78°C
Specific heat of alloy = ?
Solution:
Change in temperature:
ΔT = 78°C - 25°C
ΔT = 53°C
Now we will put the values in formula.
Q = m.c. ΔT
250 j = 25 g × c ×53°C
250 j = 1325 g.°C × c
250 j / 1325 g.°C = c
c = 0.2 j/ g.°C
Answer:
The beaker holds 307.94 mL
Explanation:
As we know that the volume that beaker hold is the volume of water that occupied by it.
For this first we have to find mass of the water in the beaker
This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.
weight of water (m) = total weight - weight of beaker
Empty weight of beaker = 25.91 g
Weight of beaker with water = 333.85 g
Weight of water = 333.85 - 25.91 = 307.94 g
Density of water = 1 g/mL
We have
Mass = Volume x density
307.94 = Volume x 1
Volume = 307.94 mL
The beaker holds 307.94 mL
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
Answer:
can you help mine please
How many molecules of chlorine are needed to react with 56.Og of iron to form Iron (III) chloride (FeCl3)?
