<span><span>N2</span><span>O5</span></span>
Explanation!
When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with
<span><span>25.92 g N<span>14.01 g N/mol N</span></span>=1.850 mol N</span>
<span><span>74.07 g O<span>16.00 g O/mol O</span></span>=4.629 mol O</span>
The ratio between these is <span>2.502 mol O/mol N</span>, which corresponds closely with <span><span>N2</span><span>O5</span></span>.
B will make little difference to reaction rate while C and D will reduce the reaction rate. So the answer is A. adding heat energy to the reactants
.
Charge of nucleus is always positive whether it is element or isotope.
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
