What are the products of photosystems ii and i in noncyclic electron flow? how is atp produced in chloroplasts?

Convert 3.30 g of copper (II) hydroxide Cu(OH)2 to molecules.

ratelena [41]

Answer:

0.18× 10²³ molecules

Explanation:

Given data:

Mass of copper hydroxide = 3.30 g

Number of molecules = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 3.30 g/97.56 g/mol

Number of moles = 0.03 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

0.03 mol × 6.022 × 10²³ molecules / 1mol

0.18× 10²³ molecules

6 0

3 years ago

Red, green and blue light all have the same frequency and wavelength. True or false

Sedaia [141]

False, they are all different because they help you know different things.

4 0

3 years ago

What is the definition of critical thinking from the book ?

Naily [24]

Answer:

Critical thinking is the intellectually disciplined process of actively and skillfully conceptualizing, applying, analyzing.

Explanation:

3 0

3 years ago

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

4 years ago

Based on solubility rules what ions in water might interfere with the analysis of calcium ions by precipitation of calcium carbo

Pavlova-9 [17]

First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:

1. Slaking of quicklime, CaO
CaO + H₂O ⇒ Ca(OH)₂

2. Precipitation
Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O

The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.

Therefore, the answer is OH⁻ ion.

6 0

3 years ago

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