Answer:
=37.83783784
Explanation:
Find the total sum of all coins,
which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100
you equation should look like this
(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.
To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
14 M x V1 = 4.20 M x 200 mL
V1 = 60 mL needed of the concentrated solution
Answer is: selenium (Se).
1) electron configuration: ₃₄Se 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp⁴.
2) ₃₃As 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp³.
3) ₃₆Kr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp⁶.
4) ₃₁Ga 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp¹.
Valence electrons of selenium are 4s²4sp⁴.
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
