The heat (Q) required to raise the temp of a substance is:<span>Q=m∗Cp∗ΔT</span><span> where m is the mass of the object (25.0g in this case), Cp is the specific heat capacity of the substance (for water Cp = 1.00cal/gC, or 4.18J/gC,
and Dt is the change in temp.
You'll have to solve this twice, once with the Cp in calories, and once with the Cp in joules.
</span><span>1380.72 Joules</span>
Answer:
Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.
Explanation:
Atom refers to the smallest constituent unit of a chemical element. Molecules refer to a group of two or more atoms that are held together due to chemical bonds.
Its air and water. if its multiple choice
Answer:
Aluminum lodide - 680 degrees
fructose- 824 degrees
potassium bromide- 2,615 degrees
calcium bromide 3, 515 degrees
lowest boiling point is 680 and highest is 3,515.
hope this helps:)
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
learn more about base dissociation constant:
brainly.com/question/9234362
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