<span>In the Bronsted-Lowry model of acids and bases, a(n) _acid____ is a hydrogen donor and a(n) _base____ is a hydrogen acceptor.</span>
Answer:
Hope this helped :) good luck! ❤️
Explanation:
A <em>coolant solution</em> is a <u><em>homogeneous </em></u>mixture because the coolant particles are not chemically combined with the water (keep their properties) and they are evenly distributed throughout the water.
Answer:
Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g
Explanation:
SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.
the balanced chemical equation is as follows
Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O
Moles of Na₂SO₃ = 
Moles of HCl = 
using mole ratio method to find limiting reagent
For sodium sulfite 
for HCl 
since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction
1 mole of Na₂SO₃ produce 1 mole of SO₂
0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂
∴ Mass of SO₂ produce = mole x molar mass of SO₂
= 0.198 x 64
= 12.672 g
Well not calculus because that has nothing, well mostly nothing to do with balancing chemical equation, so B or C. Now for me personally B is way faster, though C is sometimes faster if you get lucky the way to solve it is B
Answer:
Explanation:
From the information given:


no of moles of
= 0.01 L × 0.0010 mol/L
no of moles of
= 
no of moles of
= 0.01 L × 0.00010 mol/L
no of moles of
= 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2