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3 years ago

15

Which of the following amino acids may be considered hydrophobic at physiological ph?

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer:

Aspartic acid may be considered hydrophobic at physiological.

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If a compound with a density of 9.0 g/cm^3 occupies 37.6 in^3, what is its mass in kg?

STALIN [3.7K]

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V = m d = 14 830 g × 1 c m ³ 19.32 g = 767.6 cm³

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True or False: Anushka set out to create something to measure

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4 years ago

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What is the wavelength of electromagnetic radiation with a frequency of 9.23 × 107 Hz?

Igoryamba

Answer:

The wavelength of the given electromagnetic radiation is 3.25 m.

Explanation:

We know that,

Frequency (f) = c / λ;

where,

c - speed of light = $3\times10^{8} m/s$

λ - wavelength of the radiation.

Given,

f = $9.23 \times10^{7} Hz$ ,

From the above formula,

λ = $\frac{c}{f}$

λ = $\frac{3\times10^{8} }{9.23\times10^{7} }$

λ = 3.25m;

Therefore,

Wavelength of the given radiation = 3.25 m

3 0

4 years ago

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago