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gregori [183]
2 years ago
7

Task B: Use the following maps to complete the questions:

Chemistry
1 answer:
Oksana_A [137]2 years ago
4 0

Answer:

<h3>1. B</h3><h3>2. A</h3><h3>3. B</h3><h3>4. B</h3><h3>5. C</h3><h3>I HOPE IT HELPS :) 100% sureness</h3>
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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Why ammonium is not a molecular ion
In-s [12.5K]

Answer:

See below

Explanation:

Ammonium (NH_{4}^+) is not a molecular ion because it is just a poly-atomic ion. A molecular ion has a "negative or positive charge" as a whole but the positive charge on here is not on the whole. So, it is a poly-atomic ion and not molecular ion.

3 0
3 years ago
Which of these contains generic material but is not classified as living?
denis-greek [22]
Viruses are not classified as being alive in because of the fact that they can't reproduce.
4 0
3 years ago
When table salt (NaCl) is added to water, it lowers the freezing point is more salt is added, the freezing point decreases even
AleksAgata [21]
The answer to the question is B
7 0
3 years ago
In a percentage composition investigation a compound was decomposed into its elements: 20.0 g of calcium, 6.0 g of carbon, and 2
inysia [295]

The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

=50 g

Percentage composition :

  • Ca-calcium

\tt \dfrac{20}{50}\times 100\%=40\%

  • C-carbon

\tt \dfrac{6}{50}\times 100\%=12\%

  • O-oxygen

\tt \dfrac{24}{50}\times 100\%=48\%

3 0
2 years ago
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