1.) Write down the number of atoms that comprise each compound on either side of the equation. Using the chemical equation you can identify the atoms of each element in the reaction. Because a chemical reaction can never create or destroy new matter, a given equation is unbalanced if the number (and types) of atoms on each side of the equation don't perfectly match.Don’t forget to multiply through by a coefficient or subscript if one is present.
For example, H2SO4 + Fe ---> Fe2(SO4)3 + H2On the reactant (left) side of the equation there are 2 H, 1 S, 4 O, and 1 Fe.On the product (right) side of the equation there are 2H, 3 S, 12 O, and 2 Fe.
hi
1 mile is 1 609 meters. (rounded).
As there is 100 centimeters in a meter, then
1 mile is 160 900 centimeter.
Then 1.5 miles , is 1 mile + one half so :
160 900 + 160 900 /2 = 160 900 + 80 450
= 241 350
1.5 mile is 241 350 centimeters.
please note that the value of a mile was rounded.
He discovered this by firing Alpha particles at a gold foil and most alpha particles went through the foil and a few were deflected. The area where the alpha particles were deflected were assumed where the small dense positively charged nucleus was located. And electrons were scattered in the empty space out side the small nucleus.
Also be sure to check out my chemistry question
The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>
Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

Answer:
Specific heat of metal of the metal is 0.8394J/g°C
Explanation:
The heat the water gain is the same losing for the metal. The equation is:
m(Metal)*ΔT(Metal)*S(Metal) = m(Water)*ΔT(Water)*S(Water)
<em>Where m is mass: 66.0g water and 28.5g Metal</em>
<em>ΔT is change in temperature: (95.25°C-27.84°C) = 67.41°C for the metal and (27.84°C - 22.00°C) = 5.84°C for the water</em>
<em>And S is specific heat of water (4.184J/g°C) and the metal</em>
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Replacing:
28.5g*67.41°C*S(Metal) = 66.0g*5.84°C*4.184J/g°C
S(Metal) = 0.8394J/g°C
<h3>Specific heat of metal of the metal is 0.8394J/g°C</h3>
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