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2 years ago

Which term describes the difference in electrical charge across a membrane?

Chemistry

1 answer:

Dmitry_Shevchenko [17]2 years ago

8 0

Answer:

Resting potential

Explanation:

The resting potential is the term used to the describe the difference in electrical charge across a membrane. Ion is solution are capable of carrying a charge so their presence in biological systems is of interest. Two factors affect the resting potential:

the concentration gradients of the ions in solution across the membrane
the permeability of the said membrane to the ions mentioned above

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At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago

Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s

Triss [41]

Answer:

D. Sulfur Hexafluride

Explanation:

above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
we know that the density of water is 1.0 so the substance in order for it to float has to be less than 1.0 and the densities for Sulfer Hexa, are all less than 1!!

I hope this helped !!

4 0

2 years ago

Given the statements:

olga nikolaevna [1]

C. The salt bridge maintains the flows of ions and allows electrons to move from the anode to the cathode.

3 0

2 years ago

Read 2 more answers

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago

Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.

KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :