Answer:
Explanation:
1 mole of any substance contains the same number of particles. The particles can vary (atoms, molecules, formula units), but there are always 6.022*10²³ particles. In this case, the particles are formula units of potassium nitrate or KNO₃.
Let's create a ratio.
Since we are trying to find the formula units in 0.250 moles, we multiply by that number.
The units of moles of potassium nitrate cancel.
The denominator of 1 can be ignored, so we can make a simple multiplication problem.
If we round to the nearest tenth, the 0 in the hundredth place tells us to leave the 5 in the tenth place.
0.250 moles of potassium nitrate is approximately equal to 1.5*10²³ formula units of potassium nitrate and choice B is correct.
<span>1. D, 2. A, 3. B, 4. C, 5. D, 6. A, 7. B, 8.B....I think</span>
Barium (Ba) has the lowest electronegativity of the choices you have given.
The electronegativity of Calcium (Ca) is 1
The electronegativity of Strontium (Sr) is 0.95
The electronegativity of Magnseium (Mg) is 1.31 (has the highest)
The electronegativity of Barium (Ba) is 0.89
Of the list above, you can see that Barium has the lower level of electronegativity (0.89). Therefore, the correct answer is D.
Let me know if you need further info.
- Dotz
Answer:
There is 17.1 kJ energy required
Explanation:
Step 1: Data given
Mass of ethanol = 322.0 grams
Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K
Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K
Specific heat capacity = 2.44 J/g*K
Step 2: Calculate energy
Q = m*c*ΔT
⇒ m = the mass of ethanol= 322 grams
⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K
⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K
Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ
There is 17.1 kJ energy required
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:
The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:
Expression for is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
