Answer:
The mass of tin is 164 grams
Explanation:
Step 1: Data given
Specific heat heat of tin = 0.222 J/g°C
The initial temeprature of tin = 80.0 °C
Mass of water = 100.0 grams
The specific heat of water = 4.184 J/g°C
Initial temperature = 30.0 °C
The final temperature = 34.0 °C
Step 2: Calculate the mass of tin
Heat lost = heat gained
Qlost = -Qgained
Qtin = -Qwater
Q = m*c*ΔT
m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)
⇒with m(tin) = the mass of tin = TO BE DETERMINED
⇒with c(tin) = the specific heat of tin = 0.222J/g°C
⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C
⇒with m(water) = the mass of water = 100.0 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C
m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C
m(tin) = 163.9 grams ≈ 164 grams
The mass of tin is 164 grams
Answer:
C.
Explanation:
Butane (C4H10) has 3 C-C bonds in the carbon chain and 10 C-H bonds
Answer:
C
Explanation:
it wouldn't be A because that makes no sense
theirs no pill or any type of drug that can slow it down but their is a treatment plan for it
(It could be c or d)
Answer:
0.01917 m^3/kg.
Explanation:
Given:
P = 15 MPa
= 1.5 × 10^4 kPa
T = 350 °C
= 350 + 273
= 623 K
Molar mass of water, m = (2 × 1) + 16
= 18 g/mol
= 0.018 kg/mol
R = 0.4615 kPa·m3/kg·K
Using ideal gas equation,
P × V = n × R × T
But n = mass/molar mass
V = (R × T)/P
V/M = (R × T)/P × m
= (0.4615 × 623)/1.5 × 10^4
= 0.01917 m^3/kg.
Answer:
A. K
Step-by-step explanation:
Remember the trends in the Periodic Table:
- Atomic radii <em>decrease</em> from left to right across a Period.
- Atomic radii <em>increase</em> from top to bottom in a Group.
- Ionic radii of metal cations are <em>smaller</em> than those of their atoms.
Thus, the largest atoms are in the lower left corner of the Periodic Table.
The diagram below shows that K is closest to the lower left, so it is the largest atom. It is also larger than any of the cations.