Answer:
NOT C
Explanation:
got it wrong
Answer:
$182 is the value of the gold in the coin
Explanation:
Diameter is 2r, the ratio of the coin is = 1.62cm / 2 = 0.81cm
Following the formula of the volume of the coin:
V = π r² h
V = π*(0.81cm)²*0.085cm
V = 0.172cm³ = mL
As the density of the gold is 19.32g/mL, the mass of 0.1752mL of gold is:
0.1752mL * (19.32g / mL) =
3.385g is the mass of the coin
As the price of the gold is $53.90 / g, the value of the gold in the coin is:
3.385g * ($53.90 / g) =
To find the value of the gold in the coin, calculate its volume using the formula for the volume of a cylinder. The value of gold in the coin is $185.37.
To calculate the value of gold in the coin, we need to find its volume first using the formula for volume of a cylinder:
The radius (r) of the coin is half its diameter, so it is 0.81 cm. The height (h) is 0.085 cm. Plugging in these values, we can find the volume of the coin. Next, we calculate the mass of the gold in the coin using its density and the volume we just calculated. Finally, we multiply the mass by the price per gram of gold to find the value of the gold in the coin.
The volume of the coin is:
= 0.178 cm3.
The mass of the gold is:
m = density x volume
= 19.32 x 0.178
= 3.443 g.
The value of the gold in the coin is the value as -
= mass x price per gram
= 3.443 g x $53.90/g
= $185.37.
#SPJ3
There are two types of conductance
a) molar conductance
b) specific conductance
specific conductance is also known as conductivity
Molar conductance [the conductance offered by one mole of electrolyte dissolved in any volume of solution] increases with increase in dilution. The reason for weak and strong electrolytes are different.
i) for weak electrolytes the molar conductance increases due to increase in degree of dissociation
ii) for strong electrolytes the molar conductance increases due to decrease in inter ionic interactions
Conductivity (specific conductance) is the conductance offered by one unit volume of an electrolytic solution
The conductivity decreases with increase in dilution due to less number of ions per unit volume of solution.
so here the conductivity of resulting solution will decrease if water is added to a 0.10 M NaCl solution
important nutrients needed by fish in the aquatic
environment to survive.
Which scenarios may be explained by the facts on
the left? Check all that apply.
ID There is more dissolved oxygen in colder
waters than in warm water.
IM
Fact 2: Unlike solids, whose solubility increases
with increasing temperature, the solubility of gases
generally decreases with increasing temperature.
IND
There is less dissolved oxygen in colder
waters than in warm water.
Fish life in the ocean is more abundant during
the seaons with warmer water than seasons
with cooler water
HII
If ocean temperatures rise, then the risk to
the fish population increases.
DONE W
Answer:
A. There is more dissolved oxygen in colder waters than in warm water.
D. If ocean temperature rise, then the risk to the fish population increases.
Explanation:
Conclusion that can be drawn from the two facts stated above:
*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.
*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.
*Fishes, therefore, would thrive best in colder waters than warmer waters.
The following are scenarios that can be explained by the facts given and conclusions arrived:
A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)
D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).
Answer: the answers are A and D
Explanation:
i got it right
Explanation:
Sodium Sulphate Formula is
Na2SO4
where the ionic form is given as,
Na2+ SO42-
Answer: D. A buildup of mostly methane gas, which is pontentially explosive.
Explanation:
Firedamp refers to a gas mixture, largely methane, that appear in underground coal mines. It is explosive at concentrations between 5 and 10% in the air.