Answer:
= 374.90 kPa
Calculation:
As we know atm and kiloPascal are related to each other as,
1 atm = 101.325 kPa
So,
3.70 atm = X
Solving for X,
X = (3.70 atm × 101.325 kPa) ÷ 1 atm
X = 374.90 kPa
This is because oxygen (2.8.6) requires two electrons on its valence shell to attain stable configuration (2.8.8). Hydrogen (1) on the other hand requires one electron on its valence shell to attain stable configuration (2). Therefore in a covalent bond, it requires two hydrogen and one oxygen to share electrons and achieve stable configuration.
First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper
The change in temperature (ΔT) : 56.14 ° C
<h3>Further explanation</h3>
Given
Cereal energy = 235,000 J
mass of water = 1000 g
Required
the change in temperature (ΔT)
Solution
Heat can be formulated :
Q = m . c . ΔT
c = specific heat for water = 4.186 J / gram ° C
235000 = 1000 . 4.186 . ΔT
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
