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tiny-mole [99]
3 years ago
7

example best shows that the chemistry of water is helpful to plants? Water’s polarity produces a high density, which allows wate

r to move to the leaves. Water’s bent shape causes a slow passage of nutrients up to the leaves of plants. Water’s polarity causes cohesion that pulls other water molecules up through a plant. Water’s bent shape reduces its own passage through the cell membranes of roots.
Chemistry
2 answers:
Gennadij [26K]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "Water’s polarity produces a high density, which allows water to move to the leaves." Example best shows that the chemistry of water is helpful to plants is that <span>Water’s polarity produces a high density, which allows water to move to the leaves.</span>
faust18 [17]3 years ago
7 0

Answer:   Water’s polarity causes cohesion that pulls other water molecules up through a plant.

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Why did most of the alpha particles go straight through the gold foil in Rutherford's experiment?
andre [41]

Answer:

The fact that most alpha particles went straight through the foil is because the atom is mostly empty space.

Those that passed straight through did so because they didn't encounter any nuclei.

Explanation:

4 0
2 years ago
1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is
earnstyle [38]

3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

\frac{6}{80}=\frac{V_{2} }{40}

V_{2}=\frac{6*40}{80}=3 L

So, 3 L will be the final volume for the gas as per Charle's law.

4 0
3 years ago
Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has
bija089 [108]
<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic  \: mass }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 72 \times84.9 + 28 \times 87  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 6112.8 + 2436  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 8548.8  }{100} \\

\footnotesize \longrightarrow \:  \bf Average \:  atomic  \: mass =  85.488 \: amu  \\

8 0
2 years ago
The cost of painting the circular traffic sign shown below is $3.50 per square foot. How much, to the nearest dollar, will it co
KiRa [710]

The cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

<u>Explanation:</u>

The cost of painting the circular traffic sign is given as 3.50 dollar per square feet. The diameter of the traffic sign is 36 inch, then its radius will be \frac{36}{2} = 18 inches.

But as the cost is given in unit of feet, we have to convert the radius from inches to feet.

1 inches = 0.0833 feet

18 inches = 18 × 0.0833 feet

So, the radius of the traffic sign will be approximately equal to 1.5 feet.

The area of the traffic sign will be \pi r^{2} = 3.14 \times 1.5 \times 1.5 = 7.065 square feet.

So, if the cost of painting 1 square feet of traffic sign is 3.50 dollar, then

cost of painting 7.065 square feet of traffic sign =  3.50 × 7.065 = 24.7 dollar.

Thus, the cost of painting the traffic sign of 7.065 square feet is approximately equal to 25 dollar.

8 0
3 years ago
How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of
alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0
3 years ago
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