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3 years ago

In the diagram of the earth's interior, which part causes the diffraction of P waves made by earthquakes

Chemistry

2 answers:

I am Lyosha [343]3 years ago

6 0

Answer : The inner core and the inhomogeneous mantle.

Explanation : The inner core and the inhomogeneous mantle present in the Earth's interior, are the parts that causes the diffraction of P waves made by earthquakes. The name P-wave may have abbreviated form for either pressure wave (as it consists of compressions and rarefactions) or the primary wave (because of high velocity and first wave to be recorded by a seismograph).

Oxana [17]3 years ago

3 0

The Core and an inhomogeneous Mantle cause diffraction of P-waves. P-waves are a type of elastic seismic wave that travel through a continuum and are the first waves from an earthquake to arrive at a seismograph.

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What is the mass, in grams, of 9.01 x 1024 molecules of methanol (CH, OH) ?

Zina [86]

Answer:

480.6 g

Explanation:

Given data:

Number of molecules of methanol = 9.01 ×10²⁴

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

9.01 ×10²⁴molecules ×1 mol /6.022 × 10²³ molecules

1.5 ×10¹ mol

15 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 15 mol × 32.04 g/mol

Mass = 480.6 g

3 0

3 years ago

Do your body cells always use cellular respiration to break down glucose? Explain your answer.

Goryan [66]

Answer:

Yes cellular respiration is the only way to break down glucose. Cellular respiration takes place by the cell using oxygen to break down glucose.

7 0

3 years ago

Naturally occurring indium has two isotopes, indium-113(112.9040580 amu) and Indium-115 (114.9038780 amu. The atomic mass of ind

Kruka [31]

Answer:

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu) ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂) / 100

%₁ = x I established this is an equation

%₂ = 100 - x

Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

5 0

4 years ago

A gas that was cooled to 200 Kelvin has a volume of 65.8 L. If its initial volume was 132.4 L, what was its initial temperature?

Mandarinka [93]

Answer:

Initial temperature, T1 = 99.4 Kelvin

Explanation:

Given the following data;

Initial volume, V1 = 65.8 Litres
Final temperature, T2 = 200 Kelvin
Final volume, V2 = 132.4 Litres

To find the initial temperature (T1), we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

$\frac {V}{T} = K$

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

Making T1 as the subject formula, we have;

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

Substituting the values into the formula, we have;

$T_{1} = \frac {65.8 * 200}{132.4}$

$T_{1} = \frac {13160}{132.4}$

Initial temperature, T1 = 99.4 Kelvin

6 0

3 years ago

Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima

Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0

3 years ago