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Natasha2012 [34]
2 years ago
7

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

ll be the final temperature of the system
Chemistry
1 answer:
zlopas [31]2 years ago
6 0

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!

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How many moles of oxygen are there in 3 moles of Zn(NO3)2? (2 points)
muminat

there are 6 moles of oxygen in Zn(No3)2

4 0
2 years ago
Which of the following is true about a system at equilibrium? a The concentration(s) of the reactant(s) is equal to the concentr
anyanavicka [17]

Answer:

c The concentration(s) of reactant(s) is constant over time.  

Step-by-step explanation:

When the reaction A ⇌ B reaches equilibrium, the concentrations of reactants and products are constant over time.

a is <em>wrong</em>, because the concentrations of reactants and products are usually quite different.

b is <em>wrong</em>, because both product and reactant molecules are being formed at equilibrium.

d is <em>wrong</em>. The rates of the forward and reverse reactions are equal, but they are not zero.

6 0
3 years ago
- How much power does it take to lift a<br> 1,000 N load 10 m in 20 s?
Mariana [72]

Answer:

"500 Joule/sec" is the right answer.

Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

Now,

The power will be:

=  \frac{Force\times Velocity}{Time}

On putting the values, we get

=  \frac{1000\times 10}{20}

=  \frac{10000}{20}

=  500 \ Joule/sec

5 0
3 years ago
Hydrogen cab be obtained economically as a byproduct in the electrolysis of____________
yarga [219]
Water, because electrolysis is using electricity to break the bond of water to release 2 Hydrogens and the 1 Oxygen.
4 0
3 years ago
How many grams of ethylene glycol (c2h6o2) must be added to 1.00 kg of water to produce a solution that freezes at -5.00oc? (kf
leonid [27]

Answer: 167 g


Explanation:


1) The depression of the freezing point of a solution is a colligative property ruled by this equation:


ΔTf = i × m × Kf


Where:


ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.


i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1


Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C


2) Calculate the molality (m) of the solution


ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m


3) Calculate the number of moles from the molality definition


m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent


moles of solute = 2.69 m × 1.00 kg = 2.69 moles


4) Convert moles to grams using the molar mass


molar mass of C₂H₆O₂ = 62.07 g/mol


mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g

6 0
2 years ago
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