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slamgirl [31]
3 years ago
6

A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te

mperature of the water in the calorimeter is determined to be 22.6 °C. To this is added 50.0 mL of water that was originally at a temperature of 54.5 °C. A careful plot of the temperatures recorded after this established the temperature at T0 was 30.31 °C. What is the calorimeter constant in J/°C for this calorimeter?
Chemistry
1 answer:
Grace [21]3 years ago
5 0

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

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elena55 [62]

Answer:

According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.

Explanation:

C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)

We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.

According to Le Chatelier's principle,

1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.

2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.

3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.

4 0
3 years ago
Read 2 more answers
What is the molar mass of NH4?
astraxan [27]

Answer:

NH4

Explanation:

7 0
2 years ago
Which of the following is not a reason why actual yield is less than theoretical yield?
Gnom [1K]
Your answer is B, conservation of mass


Recall that percent yield is given by: %Yeild = actual yeild/theoretical yeild x100

During experiments, there are errors made:

• uncertainty in measurements

• losses of reactants and products

• impurity in reactants

• losses during separation (e.g. filtration or purification)

• Some side reactions might also happen.



Among the given options, only conservation of mass does not contribute to a lower actual yield compared to the theoretical yield.

5 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
2 years ago
1. This balanced equation represents a chemical reaction.
Dafna1 [17]

Answer:

combustion is the answer

5 0
2 years ago
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