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3 years ago

What are some abiotic factors that ocean creatures near the shore need to be adapted to?

Chemistry

1 answer:

alexdok [17]3 years ago

5 0

Answer:A.salinity levels, air quality, water density

Explanation:nba youngboy

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Question 1 of 10What does the second quantum number (1) describe?O A. Which sublevel the electron is inB. Which energy level is

scoundrel [369]

Explanation:

The second quantum number also called the orbital quantum number describes the type of orbital or shape of it.

Answer: D. The specific orbital within a sublevel.

7 0

1 year ago

If one pound is the same as 454 grams, then convert the mass of 78 grams to pounds.

LiRa [457]

Answer:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

If you want additional help with chemistry or another subject for FREE, check out growthinyouth.org.

4 0

3 years ago

How can you tell from the name the types of bonds present in a hydrocarbon?

Darya [45]

Answer:

Alkane

Alkene

Alkyne

Explanation:

Alkane=1 bond (Saturated hydrocarbon)

Alkene= 2 bonds (Unsaturated hydrocarbon)

Alkyne= triple bonds (Unsaturated hydrocarbon)

Formula of Alkane = CnH2n+2

Formula of Alkene = CnH2n

Formula of Alkyne = CnH2n-2

7 0

3 years ago

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

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

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

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3 years ago

What are 2 ways to give an object potential energy?

natita [175]

There are several ways to give an object potential energy. One can move the object against the force of gravity to increase. One can also stretch an object out or put pressure on it.

7 0

4 years ago