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DaniilM [7]
3 years ago
9

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

er containing 0.000010 ppm of that chemical
Chemistry
1 answer:
blsea [12.9K]3 years ago
5 0
Easy ans I don't know .......
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How do the Carnivorous plants survive without soil?
Mkey [24]
Answer:

Carnivorous plants are easy to grow, if you follow a few, simple rules.

Wet all of the time.
Mineral-free water.
Mineral-free soil.
Lots of light.


Wet all of the time.
Carnivorous plants are native to bogs and similar nutrient-poor habitats. As a consequence, the plants live in conditions that are constantly damp. To grow healthy carnivorous plants, it is important to duplicate their habitat as closely as possible. Keep the soil wet or at least damp all of the time. The easiest way to do this is use the tray method. Set the pots in a tray or saucer, and keep water in it at all times. Pitcher plants can grow in soggy soil with the water level in the saucer as deep as 1/2 the pot, but most carnivorous plants prefer damp to wet soil, so keep the water at about 1/4 inch and refill as soon as it is nearly gone. Water from below, by adding water to the tray, rather than watering the plant. This will avoid washing away the sticky muscilage of the sundews and butterworts and keep from closing the flytraps with a false alarm.


Mineral-free water.
Always use mineral-free water with your carnivorous plants, such as rainwater or distilled water. Try keeping a bucket near the downspout to collect rainwater. Distilled water can be purchased at the grocery store, but avoid bottled drinking water. There are simply too many minerals in it. The condensation line from an air conditioner or heat pump is another source of mineral-free water. Reverse-osmosis water is fine to use. Carnivorous plants grow in nutrient poor soils. The minerals from tap water can “over-fertilize” and “burn out” the plants. In a pinch, tap water will work for a short while, but flush out the minerals with generous portions of rainwater, when it is available.


Mineral-free soil.
The nutrient poor soils to which the carnivorous plants have adapted are often rich in peat and sand. This can be duplicated with a soil mixture of sphagnum peat moss and horticultural sand. Be sure to check the peat label for sphagnum moss. Other types will not work well. The sand should be clean and washed. Play box sand is great, and so is horticultural sand. Avoid “contractor’s sand” which will contain fine dust, silt, clay and other minerals. Never use beach sand or limestone based sand. The salt content will harm the plants. The ratio of the mix is not critical, 1 part peat with 1 part sand works well for most carnivorous plants. Flytraps prefer a bit more sand, and nepenthes prefer much more peat. Use plastic pots, as terra cotta pots will leach out minerals over time and stress your plants.

Explanation:


Kayo na Po bahala magpaigsi
5 0
2 years ago
How to balance NH3+NO=N2+H2O
Kryger [21]

Explanation:

NH₃+NO=N₂+H₂O;

NH₃+NO=N₂+6H₂O;

NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=5N₂+6H₂O.

3 0
1 year ago
State what would be observed when the following pairs of reagents are mixed in a test tube.
Mandarinka [93]

Answer:

(i). C6H2COOH and Na2CO3(aq)

observation: <u>Bubbles</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>colourless</u><u> </u><u>gas</u><u> </u><u>(</u><u>carbon</u><u> </u><u>dioxide</u><u> </u><u>gas</u><u>)</u>

(ii) CH3CH2CH2OH and KMnO4 /H

observation: <u>The</u><u> </u><u>orange</u><u> </u><u>solution</u><u> </u><u>turns</u><u> </u><u>green</u><u>.</u>

[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>oxidation</em><em> </em><em>of</em><em> </em><em>propanol</em><em> </em><em>to</em><em> </em><em>propanoic</em><em> </em><em>acid</em><em> </em><em>occurs</em>]

(iii) CH3CH2OH and CH3COOH + conc. H2SO4

observation: <u>A</u><u> </u><u>sweet</u><u> </u><u>fruity</u><u> </u><u>smell</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>

[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>

(iv) CH3CH = CHCH3 and Br2 /H2O

observation: <u>a</u><u> </u><u>brown</u><u> </u><u>solution</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>

4 0
2 years ago
To which group of the periodic table do lithium and potassium belong?
s344n2d4d5 [400]

Answer:

1)alkali metals

2)1g/ml

3) mercury

Explanation:

<em><u>HEY!</u></em><em><u>!</u></em><em><u>^</u></em><em><u>^</u></em>

<em><u>have </u></em><em><u>a </u></em><em><u>good </u></em><em><u>day</u></em>

<em><u>thank </u></em><em><u>me </u></em><em><u>later</u></em>

<em><u>carryonlearing </u></em>

6 0
1 year ago
Read 2 more answers
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
2 years ago
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