Using the specific heat capacity formula:
q = mc ∆ t
60.0 J = (6g)(x)(11*C)
x = 0.9 J/g*C
Aluminum
Voulme 1= 950 mL
Volume 2= ?
Temperature 1 = 25 C
Temperature 2 = 50 C
Convert your temperature to Kelvin
C+273=K
Temperature 1 = 25 C + 273 = 298 K
Temperature 2 = 50 C + 273 = 323 K
Plug in to the Formula
950 mL/298 K = ? / 323 K
Rearrange the formula to make one to solve for what is missing.
To get 323 K out of the denominator multiply by it.
Making it
950 mL x 323 K / 298 K = ?
Plug it in
950 mL x 323 K / 298 K = 1027.9 mL
Answer:
sulphuric acid as it is strong electrolyte and thus, ionises completely.
Explanation:
Answer : The expression for reaction quotient will be :
(1) ![Q_c=\frac{[SO_2][HF]^4}{[SF_4]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BSO_2%5D%5BHF%5D%5E4%7D%7B%5BSF_4%5D%7D)
(2) ![Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)
Explanation :
Reaction quotient
: It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
(1) The given balanced chemical reaction is,

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :
![Q_c=\frac{[SO_2][HF]^4}{[SF_4]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BSO_2%5D%5BHF%5D%5E4%7D%7B%5BSF_4%5D%7D)
(2) The given balanced chemical reaction is,
![2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=2MoO_2%28s%29%2BXeF_2%28g%29%5Crightarrow%202MoF%28l%29%2BXe%28g%29%2B2O_2%28g%29%5B%2Ftex%3C%2Fp%3E%3Cp%3EIn%20this%20expression%2C%20only%20gaseous%20or%20aqueous%20states%20are%20includes%20and%20pure%20liquid%20or%20solid%20states%20are%20omitted.%20%20So%2C%20the%20expression%20for%20reaction%20quotient%20will%20be%20%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DQ_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)