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pishuonlain [190]

2 years ago

9

If im 5'8 at 13 how much taller will i grow until 18?

1 answer:

fredd [130]2 years ago

6 0

The average 13-year-old height is 6’2 inches. That's an average of eight inches in three years! After 13, growth tends to slow down to about an inch every year.

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Deforestation, intensive agriculture, and drought can result in a condition called

vredina [299]

Answer:

D

Explanation:

Human activities that contribute to desertification include the expansion and intensive use of agricultural lands, poor irrigation practices, deforestation, and overgrazing. ... Some meteorologists and soil scientists measure the impacts and length of a drought to determine if it is an example of desertification.

8 0

2 years ago

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A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______

Romashka-Z-Leto [24]

To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

<span>pH = - log [H+] where H+ must be in Molar</span>

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2]

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]

[OH-] = 10^-14 / 3.25 × 10-2

[OH-] = 3.08 × 10-13 M

Answers:

Acidic

[OH-] = 3.08 <span>× 10-13 M</span>

6 0

3 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

3 years ago

For the following unbalanced equation: Cu(s) + AgNO₃(aq) ➞ Cu(NO₃)₂(aq) + Ag(s) if 1.50 moles of copper and 3.00 moles of silver

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3 years ago

Explain why it is important not to correct any gas from the first few seconds of the experiment

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Answer:

gu kha fuschhehdjdvdbeodbr

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3 years ago