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3 years ago

Which of these statements about hydrogen bonds is not true?

Chemistry

2 answers:

madreJ [45]3 years ago

8 0

A is false Hydrogen bonds have nothing to do with boiling point

just olya [345]3 years ago

5 0

B as the bonds are a cause of the high boiling temp for water

i could be wrong

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A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

<u>Step by Step by explanation:</u>

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

4 0

3 years ago

Match each term below with its definition or description.

zubka84 [21]

Answer:

1. Equivalence point

2. Direct titration

3. Primary standard

4. Titrand

5. Back titration

6. Standard solution

7. Titrant

8. Indirect titration

9. End point

10. Indicator

Explanation:

1. The equivalence point is the tiration point at which the quantity or moles of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.

2. Direct titration is a method of quantitatively determining the contents of a substance

3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance

4. A titrand is the substance of unknown concentration which is to be determined

5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration

6. A standard solution is a solution of accurately known concentration

7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution

8. Indirect titration is the process of performing a titration in athe reverse order

9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added

10 An indicator is a compound used to visually determine the pH of a solution.

5 0

3 years ago

Processes that increase the density of seawater include evaporation and _____.

Elena L [17]

Correct Answer: option C: Formation of sea ice

Reason:
<span> In cold regions, changes in salinity alters the water present in ocean. Further, water density also changes with temperature. In general, water density in ocean water increases with decreasing temperature. This is because, when salt is ejected into the ocean as sea ice forms, the water's salinity increases. Since, salt water is heavier, the density of the water increases.</span>

3 0

3 years ago

Read 2 more answers

Help plz I ONLY GOT 2 mins left

Hoochie [10]

I need more information... what was the experiment??

5 0

3 years ago

Read 2 more answers

How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The

bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz = 100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0

3 years ago