Answer:
31.9 °C
Explanation:
The formula for the heat q absorbed by an object is
q = mCΔT where ΔT = (T₂ - T₁)
Data:
q = 12.35 cal
m = 19.75 g
C = 0.125 cal°C⁻¹g⁻¹
T₂ = 37.0 °C
Calculations
(a) Calculate ΔT
q = mCΔT
12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT
12.35 = 2.406ΔT °C⁻¹
ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C
(b) Calculate T₂
ΔT = T₂ - T₁
T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C
The original temperature was 31.9 °C.
Answer:
1.Metals
These are very hard except sodium
These are malleable and ductile pieces
These are shiny
Electropositive in nature
Non-metals
These are soft except diamond
These are brittle and can break down into pieces
These are non-lustrous except iodine
Electronegative in nature
2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.
3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).
4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.
Answer:
This is confusing
but if you still need help, you can search it online
Explanation:
Answer:
Increase pressure 3X => increase Temperature 3X
Explanation:
Gay-Lussac Law => T ∝ P => T =kP => Empirical Relationship => T₁/P₁ = T₂/P₂
=> T₂ = T₁P₂/P₁
Given P₂ = 3P₁ => T₂ = T₁(3P₁)/P₁ = 3T₁
