Answer:
4.6 kJ
Explanation:
The energy required to raise the temperature of a substance from its initial temperature to a final temperature is given by:
Q = mcΔT
Where Q = energy, c = specific heat capacity of substance and ΔT = final temperature - initial temperature
Given that:
c = 2.1 kJ/kg°C, m = 87.5 = 0.0875 kg, ΔT = final temperature - initial temperature = 0 - (-25) = 25
Q = 0.0875 * 2.1 *25
Q = 4.6 kJ
Answer:
It is true. a) 0.25 mol
Explanation:
<em>Hello </em><em>there?</em>
To begin solving this problem, you have to write down the chemical equation and make sure it is well balanced.
The chemical equation is;
3Mg(s) + N2(g) => Mg3N2(s)
1 mole of Mg = 24g
We have 18g of Magnesium (Mg) reacting with Nitrogen gas (N2)
From our equation,
Mole ratio = 3 : 1, (Mg : N2)
1 mol Mg = 24g
x mol Mg = 18g
x mol Mg = (18/24) = 0.75 mol Mg
But mole ratio = 3 : 1 (Mg : N2)
This means that 3 => 0.75 mol Mg
What about ratio 1 of N2?
N2 = (0.75 mol ÷ 3)/1
= 0.25 mol N2
<em>I </em><em>hope</em><em> </em><em>this </em><em>helps</em><em> </em><em>you </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Ha</em><em>v</em><em>e </em><em>a </em><em>nice </em><em>studies.</em><em> </em><em />
<span>Mass of nitrogen = 14.0067
</span>
Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 /
14.0067 = 2.63
<span>And oxygen = 63.14 / 15.9994 = 3.95 <span>
now we have: N----- 2.63 and O----3.95
by dividing both with the smallest number we get
</span></span>
<span>N-------2.63/2.63 = 1<span>
<span>O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
</span></span></span>
N= 1 x 2 = 2
And O = 1.5 x 2= 3
<span>So, the empirical formula is N</span>₂O₃.
Answer:
Explanation:
B.
BALANCED. 7C, 16H, and 22O on each side of equation.
A.
NOT BALANCED. 7C on left and 6C on right.
C.
NOT BALANCED. 16H on left and 10H on right.
D.
NOT BALANCED. 7C on left and 14C on right.
Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of 
in 375 mL of 0.88 M of KOH = 
= 0.33 moles
Number of moles of in 496 mL of 0.76 M of CsOH = 
= 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of in mixture, ![[OH^{-}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D)
= 
Hence, ![pOH=-log[OH^{-}]=-log(0.82)=0.086](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D%3D-log%280.82%29%3D0.086)
