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3 years ago

In a genotype what does the lower case letter represent

Chemistry

2 answers:

AlladinOne [14]3 years ago

6 0

The lower case letter represents the recessive allele.

vredina [299]3 years ago

3 0

Answer:

The lower case letter represents the recessive allele.

Explanation:

I just learned abt this today so im confident that this is correct

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If you dissolve 50.0 grams of potassium bromide in 100.0g of water what is the mass percent of the resultant solution?

tatiyna

Mass percent= grams solute/ grams of solution x 100

Mass Percent= (50/ 150)x100= 33.3%

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3 years ago

Which direction does air circulate into low pressure zones in the northern and southern hemisphere

Yuri [45]

Now consider a low pressure area on a disk as shown below.A parcel of air at point A would move toward the center of the low pressure area. That movement would take it farther away from the center of the disk and therefore it would move to the west. A parcel of air at B would move toward the center of the low pressure area which would also take it closer to the center of the spinning disk where its speed is greater than the surrounding points. It would appear to move to the east. With A moving to the west and B moving to the east the line from A to B is rotating counterclockwise.

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4 years ago

Read 2 more answers

If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog

9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

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3 years ago

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$