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dexar [7]
3 years ago
8

How many grams of hydrochloric acid will react completely with a block of gold that is 3.2 cm by 3.8 cm by 2.8 cm, if the densit

y of gold is 19.3 g/ml?
Chemistry
1 answer:
VLD [36.1K]3 years ago
3 0
The chemical reaction would be:

2Au + 2HCl = 2AuCl + H2

First, we need to determine the mass of gold that is used in the reaction. We do this by multiplying the volume of the block of gold to its density. We do as follows:

Mass of Au = (3.2 cm) (3.8 cm) (2.8 cm) ( 1 mL / 1 cm^3) ( 19.3 g/ mL) = 657.1264 g Au

From the molar mass of Au which is 197 g/mol, we convert the mass of Au to moles and relate HCl by using the balanced reaction. We do as follows:

657.1264 g Au ( 1 mol / 197 g ) ( 2 mol HCl / 2 mol Au ) ( 36.46 g / mol ) = 121.62 g HCl

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A 452.3-g sample of an element at 192°C is dropped into an ice–water mixture; 110.6 g of ice melts and an ice–water mixture rema
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Answer:

the specific heat of the sample is c sam = 0.42 J/g °C

Explanation:

if we assume that the sample is submerged into the mixture, such that  all the heat released by the sample is absorbed by the ice and water, and there is no time to the heat to be released to the surroundings ( or the system is isolated):

Q ice wat  + Q sam = Q surroundings = 0

Q ice wat= - Q sam

since the ice-water mixture remains, the final temperature is 0°C (water-ice equilibrium), thus only latent heat is involved

Assuming that there is no heat due to reaction or phase change of the sample, the heat released is only sensible heat

Q sam = m sam * c sam  ( T final - T initial )

Q ice wat = m ice * ΔH fusion / M water

therefore

- m sam * c sam  ( T final - T initial ) = m ice * ΔH fusion / M water

c sam= m ice * ΔH fusion /[ m sam * M water *( T initial  - T final )]

replacing values

c sam = 110.6 g * 6.02 kJ/mol / ( 452.3 g * 18 gr/mol*( 192°C -0°C) * 1000 J/kJ = 0.42 J/g °C

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c sam = 0.42 J/g °C

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