The given question is incomplete. The complete question is :
A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
Answer: The concentration of barium acetate solution is 0.375 mol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in ml
moles of
= 
Now put all the given values in the formula of molality, we get


Therefore, the concentration of solution is 0.375 mol/L
Answer:
LHS-RHS
1 C - 1 C
4 H-2 H
2 O- 3 O
So on right side 2 hydrogen are less and one oxygen is more..so
Balanced equation is
CH4+2O2==CO2 + 2H2O
Explanation:
Well when a particle of air is becomes heated it rises, right? So you could write some like you started off close to the earth (aka the troposphere) until you became heated then you started to rise and as you reached higher elevations you cooled down and you were recycled into cool air and you moved back down and became new fresh cool air until the next time you'll become heated and rise again to be recycled into fresh cool new air.
Answer:
Explanation:
ndicator species (IS) are animals, plants, or microorganism used to monitor changes in our environment. For example, they can tell us about the impact of pollution on an ecosystem, or how well an impaired environment is being managed or restored.
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag