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jonny [76]
3 years ago
5

What is concentrated salt solution added to the tube with the cheek cells? ​

Chemistry
1 answer:
sattari [20]3 years ago
6 0

Answer:

Adding a concentrated salt solution to the solution containing cheek cells will cause the water to move outside the cells by osmosis. The cheek cells are at a higher water potential than the outside solution so, the water will move out, down the water potential gradient, until an equilibrium is reached.

Hope that answers the question, have a great day!

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The summary or ending of your experiment​
Katyanochek1 [597]

Answer:

Conclusion

Explanation:

I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.

7 0
3 years ago
P4+ __02_ P4010 <br> I need help ASAP
klasskru [66]

Answer:

P_4+5O_2\rightarrow P_4O_{10}

Explanation:

Hello!

In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:

P_4+5O_2\rightarrow P_4O_{10}

Because we have four phosphorous and ten oxygen atoms at each side.

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8 0
3 years ago
Were are electrons located
lukranit [14]
Electrons are found in the cloud that's surrounded the nucleus of an atom 
8 0
3 years ago
Read 2 more answers
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of nitrous acid and sodium hydroxide are
Bess [88]

Answer:

H+(aq) +  OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

nitrious acid = HNO3

sodium hydroxide = NaOH

Step 2: The unbalance equation

HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)

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 (Ba^2+ and Br-), look like this:

H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)

H+(aq) +  OH-(aq) → H2O(l)

7 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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