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DIA [1.3K]
3 years ago
12

Suppose 45.0 g of water at 85 °C is added to 105.0 g of ice at 0 °C. The molar heat of fusion of water is 6.01 kJ/mol, and the s

pecific heat of water is 4.18 J/g °C. On the basis of these data, (a) what will be the final temperature of the mixture and (b) how many grams of ice will melt?
Chemistry
1 answer:
Elis [28]3 years ago
4 0

Answer:

A. Final Temp = 36.428C and

47.9g ice will melt

Explanation:

Given the following data:

Mass of water (M1) = 45.0g = 0.045kg

Temperature (T1) = 85C = 358k

Mass of ice (M2) = 105g = 0.105kg

Temperature (T2) = 0c = 273k

Specific heat of water (C) = 4.18j/gC = 0.00418kj/kgc

Molar heat of fusion of water = 6.01kj/mol

Therefore, heat required (q) = MCT

M1C(T1-T2) = M2C∆T

By putting the data we have

0.045×0.00418×(358-273) = 0.105×0.00418×∆T

∆t = 0.045×0.00418×85/0.105×0.00418

∆t = 36.428C

Gram of ice that would melt would be 47.9g

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3 years ago
4.1 moles of sodium carbonate to molecules of sodium carbonate.​
docker41 [41]
<h3>Answer:</h3>

2.47 × 10^24 molecules

<h3>Explanation:</h3>

One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.

That is, 1 mole of a compound =  6.022 × 10^23 molecules

Therefore,

1 mole of Na₂CO₃ = 6.022 × 10^23 molecules

Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃

we get,

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3 0
3 years ago
Cd(s)+2HCI(aq)---- H2(g)+CdCl2(aq). what volume in liters of 0.81m HCI solution would be needed to fully react with 32.71g Cd. a
BigorU [14]

Answer:

0.718L of 0.81M HCl are required

Explanation:

Based on the reaction:

Cd(s)+2HCI(aq) → H2(g)+CdCl2(aq)

<em>1 mol of Cd reacts with 2 moles of HCl</em>

<em />

To solve this question we must, as first, find the moles of Cd. With the moles of Cd we can find the moles of HCl needed to react completely with the Cd. With the moles and the molarity we can find the volume:

<em>Moles Cd -Molar mass: 112.411g/mol-:</em>

32.71g * (1mol / 112.411g) = 0.2910 moles Cd

<em>Moles HCl:</em>

0.2910 moles Cd * (2 moles HCl / 1mol Cd) =

0.5820 moles HCl

<em>Volume:</em>

0.5820 moles HCl * (1L / 0.81moles) =

<h3>0.718L of 0.81M HCl are required</h3>
6 0
2 years ago
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