Answer:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
The reaction between methane gas and chlorine gas to form hydrogen chloride and carbon tetrachloride, all in their gaseous form can be expressed through the chemical reaction below.
CH₄ + 4Cl₂ --> 4HCl + CCl₄
Let us assume that all the involved gases behaves ideally such that each mole of the gas is equal to 22.4 L.
Through proper dimensional analysis, the volume of the produced hydrogen chloride is calculated,
V(HCl) = (1.69 mL CH₄)(1 L CH₄/ 1000 mL CH₄)(1 mol CH₄/22.4 L CH₄)(4 mols HCl/1 mol CH₄)(22.4 L HCl/1 mol HCl)(1000 mL/1 L)
V(HCl) = 6.76 mL
<em>ANSWER: 6.76 mL</em>
I would classify it as a homogeneous mixture that you can't see different particles in with a human eye but if you look through it in a microscope I would classify it as a heterogeneous mixture because you can see the different sugars and saps.Hope this helped???
Answer:
Explanation:
Hello,
In this case, ammonia dissociation is:
So the equilibrium expression:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
That in terms of the reaction extent and the initial concentration of ammonia is written as:
Thus, solving by using solver or quadratic equation we find:
Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:
![pOH=-log([OH^-])=-log(0.00133)=2.88](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.00133%29%3D2.88)
And the pH from the pOH is:
Best regards.
