Answer:
See below
Explanation:
propane mole weight = 44 gm / mole
100 gm / 44 gm / mole = 2.27 moles
From the equation, 5 times as many moles of OXYGEN (O2)are required
= 11.36 moles of oxygen
at <u>STP</u> this is 254.55 liters of O2 (because 22.4 L = one mole) and
Using oxygen as 21 percent of air means that
.21 x = 254.55 = x = <u>1212.12 liters of air required </u>
"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered
Answer:
464.1 J absorbed.
Explanation:
Given data:
Specific heat of zinc = 0.39 J/g°C
Mass of zinc = 34 g
Temperature changes = 22°C to 57°C
Energy absorbed or released = ?
Solution:
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
ΔT = 57°C - 22°C
ΔT = 35°C
Q = m.c. ΔT
Q = 34 g. 0.39 J/g°C. 35°C
Q = 464.1 J
Answer:
Group 2A — The Alkaline Earth Metals. Group 2A (or IIA) of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).
Explanation:
x = 1.01
Explanation:
Given equation:
y = 1.2345x – 0.6789
y = 0.570
Problem:
Solving for x
The variables in this equation are y and x
They can take up any value since they are variables.
Since we have been given y = 0.570
y = 1.2345x – 0.6789
To solve for x, we simply substitute for y in the equation;
since y = 0.570
0.57 = 1.2345x – 0.6789
add 0.6789 to both sides;
0.57 + 0.6789 = 1.2345x – 0.6789 + 0.6789
1.2489 = 1.2345x
Divide both sides by 1.2345
x = 1.01
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