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3 years ago

What does geotropism enable plants to do

Chemistry

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

Growing toward gravity is called geotropism. Plants also exhibit phototropism, or growing toward a light source. This response is controlled by a plant growth hormone called auxin. As shown in Figure below, auxin stimulates cells on the dark side of a plant to grow longer.

Explanation:

Hope this helps and have a great day :)

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Please someone help or tell me the answer for number 2 its a quiz grade

zubka84 [21]

Answer:

1 rearrange the equation

Explanation:

3c + 2h 2 = 2 c3h4

3c +2h2 = c3 h4

3divided by 1.5

So divide 40 by 1.5

answer is 26.7

5 0

2 years ago

117 milligrams (mg) of purified product was isolated from a chemical reaction. This experimental yield of product represents a 8

Brut [27]

Answer:

131 mg

Explanation:

Percent yield = 89%

Actual yield = 117 mg

Percent yield is given by

$\text{Percent yield}=\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\\\Rightarrow 89=\dfrac{117}{\text{Theoretical yield}}\times 100\\\Rightarrow \text{Theoretical yield}=\dfrac{117}{89}\times 100\\\Rightarrow \text{Theoretical yield}=131.46\approx 131\ \text{mg}$

The theoretical yield, for this reaction is 131 mg.

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3 years ago

Within a plant, water (H2O) and carbon dioxide (CO2) can be combined (using the energy of sunlight) to produce glucose (C6H12O6)

raketka [301]

Answer:

Reactants and products

Explanation:

The reaction can be writen inthis way

$6H_{2}O + 6CO_{2}$ ⇒ $C_{6}H_{12}O_{6} + 6O_{2}$

Here we can notice that water and carbon dioxide are combined, so they are the reactants and produce glucose and oxygen that are the products.

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3 years ago

Which energy changes are associated with a liquid boiling?

astraxan [27]

Answer:

Energy is absorbed, and potential energy increases

Explanation:

hope this helps

plz mark brainliest

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2 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

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