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3 years ago

For each situation below, state whether water will move into the cell, move out of the cell, or stay the same.

1 answer:

8 0

Water moves from an area of higher water potential (aka. "more water" in simple language) to an area of lower water potential (aka. "less water" in simple language).

For A, cells in carrots have water stored in their cytoplasm, where many soluble substances may be found (e.g. sodium ions). On the other hand, pure water has no other soluble substances other than the water molecules (I.e. H2O). Pure water will thus have a higher water potential as compared to the water in carrot cells, and so, water will move from pure water into the carrot cells via osmosis down a concentration gradient.

B. Corn syrup is water that has high concentrations of sugars, thus it is very likely to have a lower water potential than the cells of carrots. Water will move from within the cells of carrots and out to the corn syrup, down a concentration gradient.

C. The water in carrot cells will stay the same, since carrot cells have the same water potential as the surrounding solution which has the same water potential as cytoplasm.

Hope this helps! :)

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Someone plz help i need to do this assignment to pass the class

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Answer:

1)0.2

2)0.72

3)0.01

Explanation:

Formula is 1)454÷2270

2)0.6×1.2

3)8÷800

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2 years ago

Drinking energy drinks makes people more aggressive.

ch4aika [34]

Outcome (dependent) variable-drinking energy drinks

Test (independent) variable- more aggressive

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2 years ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s

Rate in terms of disappearance of $O_2$ = $-\frac{1d[O_2]}{dt}$

Rate in terms of appearance of $NO_2$ = $\frac{1d[NO_2]}{2dt}$

1. The rate of formation of $NO_2$

$-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}$

$\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s$

2. The rate of disappearance of $O_2$

$-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}$

$-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

Learn more about rate law

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https://brainly.in/question/1297322

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3 years ago

How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant

nika2105 [10]

<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.025305 \ mol \ H_2O$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0

2 years ago

You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL

pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0

3 years ago