Write the net ionic equation for the reaction that occurs when 0.103 mol HBr is added to 1.00 L of the buffer solution.

Which statement describes the energy involved in diffusion? Diffusion requires energy in all cases. Diffusion requires energy on

MArishka [77]

Answer:

Both b and d can be correct

Explanation:

Generally, diffusion does not require energy (<em>making option a wrong</em>) because it is the movement of particles from a region of high concentration to a region of low concentration hence diffusion moves particles in the direction of a concentration gradient. An example of this is the passive transport (for instance, uptake of glucose by a liver cell).

However, in some cases, when diffusion is against the concentration gradient (i.e when particles move from a region of low concentration to a region of high concentration), diffusion will require energy in a case like this (<em>making option c wrong</em>). An example of this is active transport (transport of protein called sodium-potassium pump which involves pumping of potassium into the cell and sodium out of the cell).

The explanation above shows that diffusion can require energy to move particles (in or out) of the cell through the cell membrane.

3 0

3 years ago

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what is the empirical formula of vanadium 1 oxide given that 20.38 grams of vandium combines with oxygen to form 23.58 grams of

Alex

Answer:

The empirical formula is V₂O

Empirical formula of a compound is the formula that shows the simplest whole number ratio of the atoms of elements in a given compound. Empirical formula is normally calculated when the mass of each element in a compound is known or the percentage composition by mass of each element in a compound is known.

Step by Step Explanation:

Step 1: Percentage composition of each element

Percentage composition=(mass of an element/ mass of the compound)100%

Mass of Vanadium = 20.38 g

Mass of the compound = 23.58 g

% composition of Vanadium = (20.38 g/23.58 g) 100%

= 86.43 %

Mass of Oxygen = 23.58 g -20.38 g

= 3.2 g

% composition of oxygen = (3.2/g/23.58 g) 100%

= 13.57%

Step 2: Find the number of atoms of each element in the compound

Number of atoms = percentage composition/ atomic mass

Atomic mass of Vanadium = 50.94 g/mol

Number of atoms of V = 86.43 /50.94

= 1.6967

Atomic mass of oxygen = 16 g/mol

Number of atoms of O = 13.57/16

= 0.8481

Step 3: Find the simplest ratio of atoms

Vanadium : Oxygen

1.6967 : 0.8481

= 1.6967/0.8481 : 0.8481/0.8481

= 2: 1

Whole number ratio = 2 : 1

Therefore; the empirical formula is V₂O

8 0

3 years ago

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What happens to the population size of each of the pond organisms when the perch population size is decreased? How many of the s

lana [24]

I found a presentation of Food web of a pond that will greatly connect with the above problem. http://www.eduweb.com/portfolio/earthsystems/food/foodweb4.html

Normal setting:
Species Pop. size
Blue heron Medium
Perch Medium
Bass Medium
Minnows Medium
Inverts Medium
Algae Medium

If PERCH population size decreases,
Species Pop. sizeBlue heron LowPerch LowBass MediumMinnows HighInverts HighAlgae Low

As you can see, 4 other species are affected when the Perch population size decreased. Bass is not affected.

8 0

3 years ago

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

Drupady [299]

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂

3 0

3 years ago

If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?

Hitman42 [59]

Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.

4 0

3 years ago