The volume of the textbook calculated by the student written to 3 significant figures is 992cm³
- According to this question, a student calculates the volume of his textbook and arrived at an answer of 991.880 cm³.
- To round up this digit to 3 significant figures, which means that only 3 significant numbers should appear in the answer.
- To do this, the number after the third number is 8, which means we can round up the value of 1 to 2. Hence, the volume of the textbook calculated by the student is 992cm³ in 3 s.f.
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There are many ways to test and identify metal. The easiest way is observing its color. Also how reflective it is. Other ways would be boiling point, melting point, density, or conductivity of the metal.
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Answer : The volume of liquid is 420 mL.
Explanation :
Density : The mass per unit volume of a substance is known as density.
Formula used:
As we are given:
Density of mercury = 13.5 g/mL
Mass = 12.5 pounds
First we have to convert mass of sample from pound to gram.
Conversion used:
As, 1 pound = 453.6 g
So, 12.5 pounds = 453.6 × 12.5 g = 5670 g
Now we have to calculate the volume of liquid.
Now putting all the given values in this formula, we get:
Volume = 420 mL
Therefore, the volume of liquid is 420 mL.
Answer:
case1.
The addition of acid and base leads to a change in pH of the water when adding to deionized water due to fact that acid and bases dissociated in dissolving in water. If the H+ ion increases in the water as acid addition hikes it, it will result in decreasing the pH value. The intensity of the acid also affects the dissociation of the ions.
case2
Buffers are normally formed by weak acid and its conjugate base, and adding acid to the buffer it absorbs the H+ ions so the pH will be lower and adding base or increase of OH- conjugate base resists the pH value to increase.
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
