Answer:
Ball with thingies around it
The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 
10^-5 V.
<u>Explanation:</u>
Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.
Given p = 1.47 D = 1.47 3.34 10^-30 = 4.90 10^-30.
V = 1 / (4π∈о) (p cos(θ)) / (r^2)
where p is a permanent electric dipole,
∈ο is permittivity,
r is the radius from the axis of a dipole,
V is the electric potential.
V = 1 / (4 3.14 8.85 10^-12) (4.90 10^-30 1) / (55.3 10^-9)^2
V = 1.44 10^-5 V.
Answer:
a) 0.925 mol Na2CO3 can be theoretically produced
b) 0.075 moles of the excess starting material remains
Explanation:
balaced chemical:
2 NaOH(s) + CO2(g) ↔ Na2CO3(s) + H2O(aq)
1.85n 1.00n Xn
moles theor. Na2CO3:
⇒ nNaCO3 = 1.85nNaOH * ( nNa2CO3 / 2nNaOH)
⇒ nNa2CO3 = 0.925nNa2CO3
moles of the excess:
⇒moles CO2 react = 1.85nNaOH * nCO2 / 2nNaOH = 0.925n CO2
⇒moles CO2 excess = 1.00n - 0.925n = 0.075n excess CO2
First you divide the given mass of A (the initial substance) by its molar mass to get its number of moles. Then, based on a balanced chemical reaction, you divide the number of moles by the coefficient of A, then multiply by the coefficient of the product B. Finally, multiply by the molar mass of the product B (if there is a given conversion or yield %, multiply it as well), and this gives the amount of grams of the product.
