What's a good model for isostasy is

Hammerhead Shark Process of Sexual Reproduction

I don’t even know that’s just weird

3 0

3 years ago

If a wave has a wavelength of 13 meters and a period of 0.005, what's the velocity of the wave? A. 2,600 m/s B. 260 m/s C. 1,300

Rina8888 [55]

Answer:

V = 2600ms⁻¹

Explanation:

Given parameters:

Wavelength(λ) = 13meters

Period (T) = 0.005s

Period(T) is the time it takes for a full cycle of vibration to pass through. It's unit is in seconds (s)

The Velocity of waves is expressed as:

V = fλ

Where f = frequency(s⁻¹)

Frequency of a wave is the number of waves that passes through a point per unit time

f = 1/T

Where T is the period

We can therefore express Velocity of waves as a function of period

V = λ/T

Inputing the parameters, we have:

V = 13m / 0.005s

V = 2600ms⁻¹

7 0

3 years ago

Read 2 more answers

Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?

Bogdan [553]

The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

$2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}$ .

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

$2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}$ .

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0

3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

What happens to the energy of the universe during a chemical or physical process

yulyashka [42]

Answer:

unchanged

Explanation:

In any chemical of physical process, energy is neither created nor destroyed. A process that absorbs the heat from the surroundings. ... What happens to the energy of the universe during a chemical or physical process? During any chemical or physical process, the energy of the universe remains unchanged.

5 0

3 years ago