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3 years ago

11

Light will interact with objects by making them appear opaque, translucent, or transparent. Please select the best answer from t

he choices provided T F

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

True (T)

Explanation:

By absorbing or reflecting or letting light pass through, things become opaque, translucent, or transparent.

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The Great Burdock plant’s seeds have spines on them that attach to the fur of animals that brush against it. The seed then trave

Sphinxa [80]

The answer is B. Commensalism.

<span>Commensalism is a relationship between two organisms in which only one of them has benefit, and the other one is not affected. In this example, the Great Burdock's plants spread their seeds using animals, so they benefit from this relationship. On the other hand, animals neither have benefits not are harmed from the relationship.</span>

3 0

3 years ago

Read 2 more answers

How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.

kap26 [50]

Answer:

1.88 × 10²⁴ atoms

Explanation:

Step 1: Given data

Mass of sulfur: 100 g

Step 2: Calculate the moles corresponding to 100 g of sulfur

The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:

100 g × (1 mol/32.07 g) = 3.12 mol

Step 3: Calculate the number of atoms in 3.12 moles of sulfur

We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.

3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms

7 0

3 years ago

Balancing oxidation-reduction reactions <br> Mg+ N2—>Mg3N2

BartSMP [9]

Answer:

${ \sf{3Mg_{(s)} + N_{2(g)} →Mg _{3}N_{2(s)}}}$

3 0

2 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0

3 years ago

According to the image, identify the number of neutrons in the most common isotope of aluminum.

wolverine [178]

Answer:

The number of neutron in the Aluminium Isotope is :

B. 14

Explanation:

Isotopes : These are the atoms which have same atomic number but have different mass number.

<u>This image shows the average atomic mass of Al element because it is in decimals</u>.

Atomic mass = 26.98154

(Note : mass number of single isotope can never be in decimals)

It is the average of mass of different isotopes of Al

Major Isotopes of $_{13}^{26.98154}\textrm{Al}$ are :

$_{13}^{26}\textrm{Al}$ ......atomic mass = 26
$_{13}^{27}\textrm{Al}$ .......atomic mass = 27

mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)

mass of $_{13}^{26.98154}\textrm{Al}\ \approx 27$

Now calculate the neutron in $_{13}^{27}\textrm{Al}$

Number of neutron = mass number - atomic number

= 27 - 13

Number of neutron = 14

(Atomic mass is same as mass number)

5 0

2 years ago