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snow_tiger [21]

3 years ago

8

Some transport processes use transport proteins in the plasma membrane, but do not require atp. this type of transport is known

as _____.

1 answer:

N76 [4]3 years ago

8 0

Facilitated diffusion

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This is a science question for k12

bixtya [17]

Height of wave - amplitude

Higher the pitch - higher the frequency

Louder - higher the amplitude

bottom of the wave - trough

distance from crest to crest - wave length

top of the wave - crest

6 0

2 years ago

Read 2 more answers

Based on the types of elements in the compound are there ionic or covalent? A. KCI B.NO2

Alex

B.

Explanation:

iehebrkee keen enensjsb sh sry need to get the points?

7 0

2 years ago

An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the

san4es73 [151]

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length = $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$ (i.e the ratio of the ionic radius of the cation to the ionic radius of

the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a = 664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore, the ionic radius of the anion $r^- = 312.52 \ pm$

4 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

Explain how transverse waves can be produced on a rope. Then describe how pieces of the rope move as waves pass.

andrew-mc [135]

Transverse waves can be produced on a rope by moving one end of the rope up and down.The movement causes motion in the particles that make up the rope and the rope itself becomes the medium. The particles move perpendicular to the propagation. The movement also causes crests(highest point of the wave) and troughs (lowest point of the wave) which move along the direction of propagation.

6 0

3 years ago