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3 years ago

1. Why is Earth said to be in an ice age?

Chemistry

2 answers:

vlabodo [156]3 years ago

5 0

Answer:

The variation of sunlight reaching Earth is one cause of ice ages. Over thousands of years, the amount of sunshine reaching Earth changes by quite a lot, particularly in the northern latitudes, the area near and around the North Pole.

Explanation:

I hope this helps.

noname [10]3 years ago

3 0

Answer: "The geological record appears to show that ice ages start when the continents are in positions which block or reduce the flow of warm water from the equator to the poles and thus allow ice sheets to form. The ice sheets increase Earth's reflectivity and thus reduce the absorption of solar radiation."

Explanation:

i searched it up XD

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2. What is the molarity of a HCl solution containing 1.2 grams of hydrochloric acid in 0.025 L of solution?

alisha [4.7K]

Answer:

1.3M

Explanation:

Convert from grams to moles:

molar mass of HCl = 1.01g(molar mass of H) + 35.45g(molar mass of Cl) = 36.46g HCl

1.2g HCl (1mol HCl/36.46g HCl)

= .0329 mol HCl

Molarity = mol/L (important formula for concentration)

Plug your values in:

Molarity = .0329mol/.025L

1.317M - but you used two significant figures in the question, so:

1.3M

3 0

3 years ago

Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the

Marta_Voda [28]

Answer:

The Ka is 9.11 *10^-8

Explanation:

Step 1: Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

Step 2: Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

Step 3: ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

Step 4: Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8

4 0

3 years ago

when a compound containing c h and o is completely combusted in air what reactant besides the hydrocarbon is involved in the rea

Arada [10]

When a compound containing C, H and O is completely combusted in air what reactant besides the hydrocarbon is involved in the reaction is Oxygen.

<h3>What is hydrocarbon?</h3>

A hydrocarbon is an organic molecule composed completely of hydrogen and carbon in organic chemistry. Group 14 hydrides include hydrocarbons. Hydrocarbons are often colourless and hydrophobic, with scents that are weak or exemplified by gasoline and lighter fluid. They exist in a wide range of molecular forms and phases, including gases (like methane and propane), liquids (like hexane and benzene), low melting solids (like paraffin wax and naphthalene), and polymers (such as polyethylene and polystyrene). Hydrocarbon refers to naturally occurring petroleum, natural gas, and coal, as well as their hydrocarbon derivatives and refined forms, in the fossil fuel industry. The primary source of energy on the planet is the combustion of hydrocarbons.

To learn more about hydrocarbons visit:

brainly.com/question/17578846

#SPJ4

7 0

2 years ago

3 waves are shown with a line through their center. The bottom of the first wave is labeled C. A bracket labeled D connects the

Likurg_2 [28]

Answer:Label the parts of this wave.

✔ crest

✔ amplitude

✔ trough

✔ wavelength

Explanation:

8 0

3 years ago

Read 2 more answers

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the weighted average score will be 84.1.

Find more about percents here:

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I hope it helps you!

4 0

3 years ago