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Mamont248 [21]
3 years ago
13

According to the graph, how much Potassium bromide can be dissolved in 100 g of warer at 40C?

Chemistry
2 answers:
xenn [34]3 years ago
6 0
At 40 degrees Celsius, approximately 78 grams of potassium bromide can be dissolved.
coldgirl [10]3 years ago
3 0

Answer:

77.8 g potassium bromide can be dissolved in 100 grams of water at 40°C.

Explanation:

In a given a line of KBr is straight. So ,from this we see that:

At 45 degree Celsius 80 grams of KBr is soluble in 100 grams of water.

At 90 degree Celsius 100 grams of KBr is soluble in 100 grams of water.

Slope from these point comes out to be : m

(45,80) , (90,100) :

m=\frac{100 - 80}{90 - 45}=\frac{20}{45}

According to question we want to determine the mass of KBr soluble in 100 grams of water at 40 degree Celsius. Let the mass of KBr at that temperature be y.

(40, y) , (45,80) :

Slope of the straight line as calculated above = m

m=\frac{20}{45}=\frac{80-y}{45-40}

y = 77.77777 ≈ 77.78 g

77.8 g potassium bromide can be dissolved in 100 grams of water at 40°C.

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Chemistry</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                          \displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

28.2298 mol CaF ≈ 28 mol CaF

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