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3 years ago

Which statement best describes echolocation in dolphins?

Chemistry

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

Echolocation helps dolphins find food and avoid predators.

Explanation:

Dolphins have the ability to use echolocation and their normal sight to determine some information such as the distance, speed, size and structure of objects in the water around them. This information helps dolphins find food and avoid predators in their environment.

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2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t

Anon25 [30]

Answer: The concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

Explanation:

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}$

So, $\text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M$

The given chemical equation follows:

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial: 0.1567

At eqllm: 0.1567-2x 2x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

We are given:

$K_c=9.3\times 10^{-8}$

Putting values in above equation, we get:

$9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}$

So, equilibrium concentration of hydrogen gas = $2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M$

Hence, the concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

7 0

3 years ago

Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen

OLga [1]

Answer:

5.9 kg

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ: 55.84

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

m/kg: 7.0

(a) Moles of Fe

$\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}$

(b) Moles of CO

$\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}$

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

$\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}$

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ: 32.00

2C + O₂ ⟶ 2CO

n/mol: 261

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \text{4180 g O}_{2}$

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

$\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}$

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

6 0

4 years ago

Why is pressure on Jupiter enormous?

Evgen [1.6K]

Deep under Jupiter's clouds is a huge ocean of liquid metallic hydrogen. On Earth, hydrogen is usually gas. But on Jupiter, the pressure is so great inside its atmosphere that the gas becomes liquid. As Jupiter spins, the swirling, liquid metal ocean creates the strongest magnetic field in the solar system.

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4 years ago

Which layer forms the cap rock for the oil trap?

noname [10]

Layer F from this image

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3 years ago

Read 2 more answers

The atomic mass of magnesium is the weighted average of the atomic masses of

Debora [2.8K]

Answer:

2. All the naturally occurring isotopes of Mg.

Explanation:

You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.

1. and 3. are wrong. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.

4. is wrong. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.

6 0

3 years ago