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Nataly_w [17]
2 years ago
9

contrast longitudinal and transverse waves by describing the particle movement in sound and water waves :​

Chemistry
1 answer:
Airida [17]2 years ago
3 0

Answer: In a transverse wave, the particles of the medium move perpendicular to the wave's direction of travel. Transverse waves are characterized by peaks and valleys, called crests and troughs. In a longitudinal wave, the particles of the medium move parallel to the wave's direction of travel.

Explanation:

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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0
3 years ago
The line graph shows the heights of plants grown in fertilized and unfertilized soil. Based on this information, what will most
AfilCa [17]

Answer:

  • <u><em>g) Neither plant should increase by 1 cm in height.</em></u>

Explanation:

See the graph for this question on the figure attached.

The growing of the <em>plant A</em> is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it <em>most likely </em>does not increase in height.

The growing of the <em>plant B</em> is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that <em>most likely</em> this plant will not grow on the day six or grow less than 0.5 cm.

Thus, for both plants you can say that <em>on day six, most likley, neither should increase by 1 cm in height (</em>option g).

4 0
2 years ago
Part a draw the major organic product formed when the compound shown below undergoes a reaction with an excess of etoh under aci
sweet-ann [11.9K]
An aldehyde is an organic compound containing a terminal carbonyl group (C = O). This functional group, consisting of a carbon atom bound to a hydrogen atom and an oxygen atom via double bond (the general formula: CHO) is called the aldehyde group. In a reaction of the addition of alcohol to the carbonyl group, it forms hemiacetals. 
On the picture attached it is shown the reaction of alcohol addition to the carbonyl group with the major organic product <span>formed in the reaction.</span>

7 0
2 years ago
There are only two isotopes of gallium: ^69Ga (68.926 amu) and ^71Ga(70.025 amu). The average atomic mass of gallium is 69.723 a
SVEN [57.7K]
The question ask for the percentage of the abundance of galium-69 where there is two isotopes of galium: the 69Ga and the 71Ga. The average atomic mass of gallium is 69.723 amu. So the formula would be <span>69.723amu=(%x)∗(68.926amu)+(1−%x)∗(70.025amu) and the answer to this is 1.58%</span>
3 0
2 years ago
Will give most brainly.
Anna [14]

Answer:

One mole

Explanation:

It tells u the concentration

8 0
2 years ago
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