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3 years ago

Discuss how the actions of the forklifts are related to actions in cells

1 answer:

7 0

so i have to discuss how forlifts are related to actions in the cell. ok easy Answer: Forklifts collect material (protein) and dump it in storage areas. ... A Cytoplasm is like a Fork Lift because they both transport particles or substances around the factory or the Animal. Therefore, forklift and cytoplasm share this analogy of collecting materials and dumping it elsewhere.

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The electronegativity of nitrogen (N) is 3.0, while the electronegativity of hydrogen (H) is 2.1. Knowing this, consider how the

Ede4ka [16]

Answer:

NH3 is polar due to the bonds between nitrogen and hydrogen which have different electronegativity and due also to its asymmetrical shape.

Explanation:

NH3 is polar as there are 3 dipoles in the ammonia molecule that do not balance each other out.

Considering the N-H bond which is polar because N with an electronegativy of 3.0, is more electronegative than H, with an electronegativity of 2.1. The is overall asymmetrical shape of NH3

means that the dipoles remains unbalanced and do cancel out each other making the NH3 polar.

4 0

3 years ago

"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail place

Amanda [17]

Answer:

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

6 0

4 years ago

Read 2 more answers

What is the term that describe two species living together

Sergeu [11.5K]

I think it is <span>Symbiosis. But I dont know how to explain it</span>

3 0

3 years ago

Read 2 more answers

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

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

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago

Good chemistry question

NeX [460]

Compared to the charge and mass of a proton an electron has......

A proton has approximately the same mass as..........

5 0

3 years ago