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Natali [406]
3 years ago
7

Every cell is surrounded by a thin membrane. What is the main function of this cell membrane?

Chemistry
2 answers:
zubka84 [21]3 years ago
8 0

Answer:

The main function of the cell membrane is to protect the cell from the outer environment.

attashe74 [19]3 years ago
5 0

Answer: The answer is D.) to separate the inside of the cell from the outside environment

Explanation:

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which term is used when sugar is completely dissolved in water                                                                 
GaryK [48]
The term which is used is homogeneous.
when sugar is completely dissolved in the water, the mixture or solution homogeneous, both in same phase and same uniform texture that is liquid.
There two types of mixtures are heterogeneous and homogeneous in different phases.
If sugar is not completely dissolved in water and you see the crystals of sugar in water, then the solution will be heterogeneous.
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Read the given equation:
Assoli18 [71]

Answer:

D.

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Why do we call water the univeral salivnet
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3 years ago
Imagine that instead of balls, you were rolling positively charged ions of the same element).
Ghella [55]
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8 0
2 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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