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stira [4]
3 years ago
12

What is something Kierra can do to get more accurate data on her lettuce plants?

Chemistry
1 answer:
AveGali [126]3 years ago
6 0

Answer:

B

Explanation:

Planting several seeds from each brand being tested is a way of repeating the test to gain better results.

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Taxol is a potent chemotherapeutic agent (isolated from the Pacific Yew tree) which is especially effective against ovarian canc
butalik [34]

Answer:

Amine

Explanation:

The functional groups contained in Taxol are :

Ketone , Ester, Amide and Alcohol

while the functional group that is not contained in the Taxol is Amine

Taxol is a very potent anti-cancer chemotherapeutic, and it is also groped into a class called Taxanes and this makes it effective in the treatment of breast and ovarian cancer.

8 0
3 years ago
A 0.532 mol sample of SO2 gas requires 52.3 s to effuse through a tiny hole. Under the same conditions, how long will it take 0.
scoundrel [369]

Answer:

41.3 s

Explanation:

Let t₁ represent the time taken for SO₂ to effuse.

Let t₂ represent the time taken for Ar to effuse.

Let M₁ represent the molar mass of SO₂

Let M₂ represent the molar mass of Ar

From the question given above,

Time taken (t₁) for SO₂ = 52.3 s

Time taken (t₂) for Ar =?

Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol

Molar mass (M₂) of Ar = 40 g/mol

Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:

t₂ / t₁ = √(M₂ / M₁)

t₂ / 52.3 = √(40 / 64)

t₂ / 52.3 = √0.625

t₂ / 52.3 = 0.79

Cross multiply

t₂ = 52.3 × 0.79

t₂ = 41.3 s

Thus, the time taken for the amount of Ar to effuse is 41.3 s

3 0
3 years ago
Find the mass in kilograms of the liquid air that is required to produce 600L of oxygen. In normal condition, 1L of liquid air h
IgorC [24]

Mass in kilograms of liquid air required = 0.78 kg

<u>Given that </u>

1 Litre of liquid air contains 1.3 grams of oxygen ( air )

<u />

<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>

volume of air given = 600 L

mass of liquid air required = x

1 litre = 1.3 grams

600 L =  x

∴ x ( mass of liquid air ) = 1.3 * 600

                                       = 780 g  = 0.78 kg

Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg

Learn more about liquid air : brainly.com/question/636295

3 0
2 years ago
Read 2 more answers
Thank chu if you answer
Vikentia [17]

Answer: H

Explanation:

4 0
3 years ago
Which is the most concentratic acid​
denis23 [38]

Answer:

Tetraoxosulphate vi acid (H2SO4)

Explanation:

Becauses it ionizes completely

5 0
3 years ago
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