Answer:
A=covalent B=low melting point
Explanation:
just did this on edge... not sure if it is still needed but there wasn't an answer so :)
Answer:
1 = Q = 7315 j
2 =Q = -21937.5 j
Explanation:
Given data:
Mass of water = 50 g
Initial temperature = 20°C
Final temperature = 55°C
Energy required to change the temperature = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 j/g.°C.
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 = T2 - T1
ΔT = 55°C - 20°C
ΔT = 35°C
Q = 50 g× 4.18 j/g.°C×35°C
Q = 7315 j
Q 2:
Given data:
Mass of metal = 100 g
Initial temperature = 1000°C
Final temperature = 25°C
Energy released = ?
Specific heat capacity = 0.225 j/g.°C
Solution:
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 = T2 - T1
ΔT = 25°C - 1000°C
ΔT = -975°C
Now we will put the values in formula.
Q = 100 g × 0.225 j/g.°C × -975°C
Q = -21937.5 j
Negative sign show that energy is released.
Answer:
The bombarding particle is a Proton
Explanation:
A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.
¹⁶₈O(P,alpha) ¹³₇N, can be written as
¹⁶₈O + x goes to ¹³₇N + ⁴₂He
Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number
16 + x = 13 + 4
x = 17 – 16 = 1, Hence we can say that x = ¹₁P
<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>
Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle
The concentration of carbon monoxide at equilibrium is 0.209 M.
<h3>Concentration of each gas</h3>
CO2 = 2 mol/5 L = 0.4
H2 = 1.5 mol/5L = 0.3
<h3>ICE table</h3>
Create ICE table as shown below
CO2(g) + H2(g) ↔ CO(g) + H2O(g)
I 0.4 0.3 0 0
C - x - x x x
E 0.4 - x 0.3 - x x x
Kc = [CO][H₂O] / [CO₂][H₂]
2.5 = (x²)/(0.4 - x)(0.3 - x)
x² = 2.5(0.4 - x)(0.3 - x)
x² = 2.5(0.12 - 0.7x + x²)
x² = 0.3 - 1.75x + 2.5x²
0 = 1.5x² - 1.75x + 0.3
solve the quadratic equation using formula method;
x = 0.209
Thus, the concentration of carbon monoxide at equilibrium is 0.209 M.
Learn more about concentration here: brainly.com/question/17206790
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