Answer:
-195.8 °C = 77.35 K = -320.44 °F
Explanation:
Temperature is commonly measured in following three scales;
1) Kelvin
2) Fahrenheit<span>
3) </span>Celsius<span>
Other scales are Rankine, Romer, Newton, Delisle e.t.c
Temperature given is -195.8 </span>°C. Degree Celsius is related to Kelvin and Fahrenheit as follow,
Celcius to Kelvin;
K = °C + 273.15
So,
K = -195.8 + 273.15
K = 77.35
Celcius to Fahrenheit;
°F = °C × 9/5 + 32
So,
°F = -195.8 × 9/5 + 32
°F = -320.44
<span>A. Helium </span>atomic number 2
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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 :
(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)
Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
