Answer:
A. The equilibrium constant is very large
Explanation:
The equilibrium constant value is the ratio of the concentrations of the products over the reactants. When a chemical reaction goes to completion, that means that all the reactant has turned into products. As the equilibrium constant defines, it is the ratio of the product to the reactant. So at the final stage of the chemical reaction, the equilibrium constant will be very large.
A colored line, as long as it is one single piece, not broken
Answer:
B. accepted value x 0.1
Explanation:
in the equation provided
Maximum allowed value of percentage error = 10%
put this value in the equation in stead of percentage error we get,
so maximum error = .1 x accepted value
10 % percentage error means the experimental value has 10 % error compared to accepted value.so error will be 10 % of the accepted value
or .1 times of accepted value
The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
Learn more about heat transfer:
brainly.com/question/6363778
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0.250 mol/L
<em>Step 1</em>. Write the chemical equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
<em>Step 2</em>. Calculate the moles of H2SO4
Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)
= 6.25 mmol H2SO4
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)
= 12.5 mmol NaOH
<em>Step 4</em>. Calculate the concentration of the NaOH
[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L
