The third answer because there are two of each atom
Answer:
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.
What is the specific heat of the substance?
If it is one of the substances found in Table 8.1.1, what is its likely identity?
Lower flammable limit means the lowest concentration of a material that will propagate a flame.
What is hazardous atmosphere?
It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes
- Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
- Airborne combustible dust at concentration that meets or exceeds its LFL
What is lower flammable limit?
- It means the lowest concentration of a material that will propagate a flame.
- The LFL is usually expressed as percent by volume of material in air (or other oxidant)
- Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
- However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn
Learn more about lower flammable limit at brainly.com/question/2456135
#SPJ4
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
#SPJ4
The answer is A. As the first statement is a true statement. Hope this help you
