You are performing an experiment in your lab. To compare with other experiments you need your results to be in moles. During you

The thermal energy from the soup is transferred to Greg's hands. 

The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.

Answer:

B, D

Explanation:

We need to recognize that the ice will rise in temperature from -6.5 ºC to 0 ºC for it to change into water.

Let's define q₁ as the heat needed to warm the ice to 0ºC, and q₂ as the heat for the transition from solid to liquid.

The calculation for q₁ is as follows:

q₁ = s x m x ΔT, where s represents the specific heat of ice (2.09 J/gºC), m is the mass, and ΔT is the temperature difference.

For q₂, the enthalpy of fusion is computed as:

q₂ = C x ΔT

with C indicating the specific heat for the phase transition, denoted as AH in kJ/mol.

All necessary data for computing q₁, q₂, and the total heat change (q₁ + q₂) is provided.

q₁ = 25.0 g x (2.09 J/gºC) x (0 - (-6.5 ºC))

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

Combining these values gives us qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ.

Now we can answer the question:

(a) False, AH refers to the heat capacity during melting.

(b) True, as we concluded earlier.

(c) False, there’s only one phase transition from solid (ice) to liquid.

(d) True based on our calculations above.

(e) False, according to our findings.

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %

1.09 moles of tin (Sn)

Explanation:

The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):

SnCl₄ + 4 Na → Sn + 4 NaCl

Considering the chemical equation, this results in:

If 1 mole of SnCl₂ reacts with 4 moles of Na,

then 2.34 moles of SnCl₂ will react with X moles of Na.

X = (2.34 × 4) / 1 = 9.36 moles of Na

So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:

If 4 moles of Na yield 1 mole of Sn,

then 4.36 moles of Na will produce Y moles of Sn.

Y = (4.36 × 1) / 4 = 1.09 moles of Sn

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limiting reactant