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4 years ago

The atomic number of oxygen, 8 indicates that there are eight what???

2 answers:

6 0

The atomic number organizes the elements and also tells you how many protons there are in an element. So, oxygen would have 8 protons. The atomic number of oxygen 8 indicates that there are eight protons.

Hope this helps! (If correct rank as brainliest answer :)

Strike441 [17]4 years ago

4 0

The answer is A. The atomic number of an element is the number of protons in its nucleus.
Hope this helps

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Temperature of a substance is proportional to the average kinetic energy of the motion of the molecules of that substance.

MrRissso [65]

True, according to the Kinetic Molecular Theory.

4 0

3 years ago

Read 2 more answers

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!

poizon [28]

Answer:

391.67Hz

Explanation:

The fundamental frequency formula in string is expressed as;

Fo = V/2L

V is the velocity of the wave = 329m/s

L is the length of the string = 42cm = 0.42m

Substitute

Fo = 329/2(0.42)

Fo = 329/0.84

Fo = 391.67Hertz

Hence the fundamental frequency of a mandolin string is 391.67Hz

4 0

3 years ago

The mass of a glass beaker is known to be 24.2 g. Approximately 5 mL of water are added, and the mass of the beaker and water is

Kamila [148]

Answer: 2 significant figures in 6.4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.

All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.

All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.

All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.

Mass of beaker = 24.2 g

Mass of beaker + mass of water = 30.625 g

Mass of water = 30.625 - Mass of beaker = 30.625 - 24.2 = 6.4g

The rule apply for the addition and subtraction is :

The answer would contain same number of decimal places as there are in the least precise number present.

Thus there are 2 significant figures in mass of water which is 6.4 grams.

7 0

3 years ago

When atoms form a bond, it is usually because each atom acquires a more stable arrangement of electrons in its electron shells.

liraira [26]

C. Usually when an atom loses or gains an electron, it is because it is trying to satisfy the Octet Rule. The Octet Rule states that an atom is at its stablest when it has 8 valence electrons (two in helium's case)

If you look on the periodic table, elements on the left (Alkaline Metals) are the most reactive because they only have one valence electron (or electron in the outer shell). Elements on the right (Noble Gases) are the least reactive because they have a full outer shell of 8 valence electrons.

Later on you will find that as stability decreases as you go down the periodic table but that is a discussion for a different time.

4 0

3 years ago

Read 2 more answers

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

3 years ago