Answer:
the male carries the babies
the male guards the nest
The male oxygenates the embryos
Explanation: males of lump fish dont do good as pets but u can train them brrrrrrr
Answer:
680 g/m is the molar mass for the unknown, non electrolyte, compound.
Explanation:
Let's apply the formula for osmotic pressure
π = Molarity . R . T
T = T° absolute (in K)
R = Universal constant gases
π = Pressure
Molarity = mol/L
As units of R are L.atm/mol.K, we have to convert the mmHg to atm
760 mmHg is 1 atm
28.1 mmHg is (28.1 .1)/760 = 0.0369 atm
0.0369 atm = M . 0.082 L.atm/mol.K . 293K
(0.0369 atm / 0.082 mol.K/L.atm . 293K) = M
0.0015 mol/L = Molarity
This data means the mol of solute in 1L, but we have 100mL so
Molarity . volume = mol
0.0015 mol/L . 0.1L = 1.5x10⁻⁴ mole
The molar mass will be: 0.102g / 1.5x10⁻⁴ m = 680 g/m
Answer:
Volume of dry gas at STP = 0.432 liters or 432 ml
Explanation:
Given:
Pressure (P) = 740 mmHg - 24 mmHg = 716 mmHg
Temperature (t) = 25 degrees C + 273 K = 298 K
500 ml = 0.5 l
Find:
Volume of dry gas at STP
Computation:
[P1][V1] / T1 = [P2][V2] / T2
[716][0.5] / 298 K = [760][ x Liters] / 273 K
x = 0.432 Liters
Volume of dry gas at STP = 0.432 liters or 432 ml
Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
Answer:
Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.
The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.
Multiply on the right side of the equation. The units of degrees Celsius cancel.
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g
The units of Joules cancel.
The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.
The mass of the sample of metal is approximately <u>333 grams.</u>
