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Tasya [4]
3 years ago
7

An example of a secondary sex characteristic is the increase in hair on some parts of the body.

Chemistry
2 answers:
Igoryamba3 years ago
7 0

Answer:

B and E

Explanation:

Hope this helps my dude’s!

nignag [31]3 years ago
6 0

Answer:

The ovaries the testes

Explanation:

Do good!

You might be interested in
21.If the anion in an ionic compound is a nonmetal, you must change its ending
True [87]

Answer:

-ite --> Anionite

Explanation:

Feel free to give brainliest.

Have an awesome day!

7 0
3 years ago
The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr - Cp*mu/k where Cp is the heat
algol [13]

Answer:

The Prandtl number for this example is 14,553.

Explanation:

The Prandlt number is defined as:

Pr=\frac{C_{p}*\mu}{k}

To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.

\mu=1896 \frac{lbm}{ft*h}*\frac{1000 g}{2.205 lbm}*\frac{3.281 ft}{1 m}*\frac{1h}{3600s}  =7938 \frac{g}{m*s}

Now that we have coherent units, we can calculate Pr

Pr=\frac{C_{p}*\mu}{k}=0.66*7938/0.36=14553

8 0
3 years ago
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
What is the approximate mass of 72 cm3 of silver, if the density is 10.5 g/cm3?
77julia77 [94]

Answer:

<h3>The answer is 756 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of silver = 72 cm³

density = 10.5 g/cm³

We have

mass = 10.5 × 72

We have the final answer as

<h3>756 g</h3>

Hope this helps you

5 0
3 years ago
Read 2 more answers
The half-life of Pa-234 is 6.75 hours. If a sample of Pa-234 contains 112.0 g, 1 point
umka2103 [35]

Answer:

28g remain after 13.5 hours

Explanation:

Element decayment follows first order kinetics law:

ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>

<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>

Half-life formula is:

t_{1/2} =  \frac{ln2}{k}

6.75 = ln2 / k

<em>k = 0.1027hours⁻¹</em>

Using rate constant in (1):

ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]

ln[Pa-234] = 3.332

[Pa-234] = <em>28g after 13.5 hours</em>

<em />

5 0
3 years ago
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