Answer:
we will take a 100g sample of this solution for our convenience
so , there is 15 g kBr in this 100g solution
we know that molality is the number if moles of solute / mass of solvent in kg
we need to find the number of moles in 15g kBr
no of moles = 15/119 s
moles = 0.126 moles/ 100g
multiplying both the numerator and the denominator by 10 to get 1 kg in denominator
= 1.26 moles / 1 kg
Hence, the molality is 1.26
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The answer is: [D]: " 417 cm³ " .
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Explanation: Use the formula:
V₁ /T₁= V₂ /T₂ ;
in which: V₁ = initial volume = 556 cm³ ;
T₁ = initial temperature = 278 K ;
V₂ = final ("new") temperature = 308 K
T₂ = final ("new:) volume = ?
Solve for "V₂" ;
Since: V₁ /T₁= V₂ /T₂ ;
We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
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V₁ /T₁= V₂ /T₂ ; Multiply EACH side of the equation by "T₂ " :
→ T₂ (V₁ /T₁) = T₂ (V₂ /T₂) ;
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to get:
↔ T₂ (V₂ /T₂) = T₂ (V₁ /T₁) ;
→ V₂ = T₂ (V₁ /T₁) ;
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Now, plug in our known values, to solve for "V₂" ;
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→ V₂ = T₂ (V₁ /T₁) ;
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→ V₂ = 308 K ( 556 cm³ /278 K) ;
→ The units of "K" cancel to "1" ; and we have:
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→ V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note: We will keep the units of volume as: "cm³ "; since all the answer choices given are in units of: "cm³ " ; {that is, "cubic centimeters"}.
→ [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
→ For the "(115,648)" ; round to "3 (three significant figures)" ;
→ "(115,648)" → rounds to: "116,000" ;
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→ (116,000) / (278) = 417.2661870503597122 ;
→ round to 3 significant figures; → "417 cm³ " ;
→ which corresponds with "choice [D]".
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The answer is: [D]: "417 cm³ " .
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Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
Answer:
ionic bonding or covalent
Explanation:
