Answer: An amount of 
heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Explanation:
Given: mass of lead = 4.64 kg
Convert kg into grams as follows.
The standard value of specific heat of lead is 
.
Formula used to calculate heat is as follows.
where,
q = heat energy
m = mass of substance
C = specific heat of substance
= change in temperature
Substitute the value into above formula as follows.
Thus, we can conclude that heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Answer:
2Mg + O₂ ⟶ 2MgO
Explanation:
Step 1. Start with the most complicated-looking formula (O₂?).
Put a 1 in front of it.
Mg + 1O₂ ⟶ MgO
Step 2. Balance O.
We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.
Mg + 1O₂ ⟶ 2MgO
Step 3. Balance Mg.
We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.
2Mg + 1O₂ ⟶ 2MgO
Every formula now has a coefficient. The equation should be balanced. Let’s check.
<u>Atom</u> <u>On the left</u> <u>On the righ</u>t
Mg 2 2
O 2 2
All atoms are balanced.
The balanced equation is
2Mg + O₂ ⟶ 2MgO
Answer:
100
Explanation:
Step 1: Subtract 50 by 300 to find the total mass of the marbles
300 - 50 = 250
Not hard, right?
Step 2: Divide 250 by 2.5
250 ÷ 2.5 = 100
And... 100 is your answer :)
Hope this helps :)
-jp524
Ur moms the density of mercurio
