How do populations survive when the environment changes?

Chemistry

2 answers:

kkurt [141]2 years ago

6 0

With adaptation, or a mutation that provides a better way for the organism to survive in the new environment

katrin [286]2 years ago

3 0

Answer:

adaptation?

Explanation:

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MrRissso [65]

Answer:

24.9%

Explanation:

According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:

0.0927 × 100 = 9.27%

Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;

100% - 9.27% = 90.73%

THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles

N.B: mole = mass/molar mass

Given the Molar Mass

NaCl: 58.44 g/mol

H2O: 18.016 g/mol

For NaCl;

0.0927 = mass/58.44

mass = 0.0927 × 58.44

5.42g

For H2O;

9.073 = mass/18.016

mass = 9.073 × 18.016

= 16.35g

Total mass of solution = 16.35g + 5.42g = 21.77g

Mass percent of NaCl = mass of NaCl/total mass × 100

% mass of NaCl = 5.42g/21.77g × 100

= 0.249 × 100

= 24.9%

5 0

2 years ago

Read 2 more answers

4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration

anzhelika [568]

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

$percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100$