Answer: Yes
Explanation: Silicon is never found as a free element naturally, meaning it always occurs with other elements.
Answer:
Five Laboratory Safety Rules:
1). Do not eat in the laboratory.
2). Do not touch any chemical or reagent unless you are told to do so.
3). Neither play in lab, nor sit on the table.
4). Don't remove labels on any reagent.
5). Don't taste anything in the laboratory, no matter how familiar it appears.
Hope it helps.
<h3>
Answer:</h3>
134 atm
<h3>
Explanation:</h3>
- Based on the pressure law, the pressure of a gas varies directly proportionally to the absolute temperature at a constant volume.
- Therefore; we are going to use the equation;

In this case;
Initial pressure, P1 = 144 atm
Initial temperature, T1 (48°C) = 321 K
Final temperature, T2 (25°C) = 298 K
We need to find the final pressure,
Therefore;
P2 = (P1/T1)T2
= (144/321)× 298 K
= 133.68 atm
= 134 atm
Therefore, the new pressure will be 134 atm.
Answer:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
We can calculate how long the decay by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span><span>From the half-life data, we can calculate for k.
</span>
1/2(Ao) = Ao e^-k(30)
<span>k = 0.023
</span>
0.04Ao = Ao e^0.023(t)
<span>t = 140 sec</span>