Answer:1.b 2.A (i think it might be c) 3 (i have no clue) 4.(no clue) Sorry i only got 1 and 2 witch i think are right its been a while sense i have done this....

Explanation:

3 0

3 years ago

If a compound has a log kow value of 6.5, what would be its predicted concentration (in ppm) in the fat of fish that swim in wat

blsea [12.9K]

Easy ans I don't know .......

5 0

4 years ago

What is the concentration (in ppm) of Mn7+ in a 0.00300 M KMnO4 solution? Note: for dilute aqueous solutions, 1 ppm = 1 mg/L.

aleksklad [387]

We know that there is 1 mole of Mn for every 1 mole of KMnO4, therefore the molarity of Mn is similar with KMnO4:

Mn = 0.00300 M

Molar mass of Mn is 54.94 g / mol. Molarity (M) is moles / L, therefore:

Mn = (0.00300 moles / L) * (54.94 g / mol) * (1000 mg / g)

3 0

3 years ago

The block in this illustration is floating in water, which has a density of 1.00 g/cm3 .What is a good estimate of the density o

miss Akunina [59]

Answer : The density of the block is estimated to be, $0.30g/cm^3$ .

Explanation :

Density of the water, $\rho_w=1.0 g/cm^3$

As we know that,

Buoyant force = $\rho\times volume\times g$

where,

g = gravitational acceleration

volume of the block = v

density of the block = $\rho_o$

Weight of the Block = Buoyant force exerted by water

$\rho_o\times v\times g=\rho_w\times 30\%\text{ of volume of block}\times g$

$\rho_o\times v\times g=\rho_w\times 0.30v\times g$

$\rho_o\times v\times 9.8 m/s^2=1 g/cm^3\times 0.30v\times 9.8 m/s^2$

$\rho_o=0.30 g/cm^3$

Thus, the density of the block is estimated to be, $0.30g/cm^3$ .

6 0

3 years ago