Answer:
m AgCl = 28.395 g
Explanation:
- AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
∴ [AgNO3] = 0.156 mol/L
∴ V = 1.27L
⇒ mol AgNO3 = 0.156 mol/L * 1.27 L = 0.19812 mol AgNO3
mass AgCL:
⇒ m AgCl = 0.19812 mol AgNO3 * mol AgCl/molAgNO3 * 143.32gAgCl/molAgCl
⇒ m AgCl = 28.395 g
Answer:
im not doing your whole homework. but ill give you one cent ;>, whats ur paypal
Explanation:
Answer:
table sugar, C12H22O11ation
Explain:
we learned this last year
The required volume of water to make the dilute solution of 0.5 M is 188 mL.
<h3>How do we calculate the required volume?</h3>
Required volume of water to dilute the stock solution will be calculated by using the below equation as:
M₁V₁ = M₂V₂, where
- M₁ & V₁ are the molarity and volume of stock solution.
- M₂ & V₂ are the molarity and volume of dilute solution.
On putting values from the question to the above equation, we get
V₂ = (2)(47) / (0.5) = 188mL
Hence required volume of water is 188 mL.
To know more about volume & concentration, visit the below link:
brainly.com/question/7208546
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Answer: D. an open series circuit
