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3 years ago

Which is the most important part of the food web? Producers, decomposers, carnivores, or herbivores

Chemistry

2 answers:

mash [69]3 years ago

8 0

Answer:

all is important

Explanation:

goldfiish [28.3K]3 years ago

4 0

Answer:

All parts of the food chain are very important for if one is part of the food chain is tooken away than other parts of the food chain will be affected.

Explanation:

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Use the periodic table to determine the electron configuration for dysprosium (Dy) and americium (Am) in noble-gas notation

sergij07 [2.7K]

Answer:

Dysprosium [Dy]= $[Xe]4f^{10}6s^2$

Americium [Am]= $[Rn]5f^77s^2$

Dysprosium is a chemical element with symbol Dy and atomic number of 66. It is a rare earth metal and as it contains partially filled f sub shells, it belongs to f block. Xe is the nearest noble gas and has atomic number of 54.

Americium is a chemical element with symbol Am and atomic number of 95. It is a rare earth metal and as it contains half filled f sub shells, it belongs to f block. Radon is the nearest noble gas and has atomic number of 86.

3 0

3 years ago

Read 2 more answers

18 g of argon occupy 750 ml at a particular temperature and pressure. How many grams of methane would occupy the same volume at

frozen [14]

Answer:

7.21 grams is the mass of methane

Explanation:

We may use the Ideal Gases Equation to solve this:

P. V = n. R. T

Let's determine the moles of Ar

18 g . 1 mol/ 39.9 g = 0.451 mol

In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.

Let's convert the moles to mass of CH4.

0.451 mol . 16g/1mol = 7.21 grams

5 0

3 years ago

A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added

Ulleksa [173]

Answer: The pH change of the buffer is 0.30

Explanation:

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})$ .....(1)

We are given:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.306M$

$[C_6H_5NH_2]=0.418M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{initial}=14-8.99=5.01$

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-$

Initial: 0.418 0.124 0.306

Final: 0.294 - 0.430

Calculating the pOH by using using equation 1:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.430M$

$[C_6H_5NH_2]=0.294M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{final}=14-9.29=4.71$

Calculating the pH change of the solution:

$\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30$

Hence, the pH change of the buffer is 0.30

8 0

3 years ago

An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize

oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0

3 years ago

Why isn't there a lunar eclipse every time Earth is in between the sun and the Moon?

DerKrebs [107]

Because the Earth's orbit around the sun is not in the same plane as the Moon's orbit around the Earth.

8 0

3 years ago