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3 years ago

How does an area’s climate determined

Chemistry

1 answer:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

The simplest way to describe climate is to look at average temperature and precipitation over time. Other useful elements for describing climate include the type and the timing of precipitation, amount of sunshine, average wind speeds and directions, number of days above freezing, weather extremes, and local geography.

Explanation:

hope it helps

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What does literal mean in science please hurry it's due at 11:40!!

Leokris [45]

Answer:

In computer science, a literal is a notation for representing a fixed value in source code. An anonymous function is a literal for the function type. In contrast to literals, variables or constants are symbols that can take on one of a class of fixed values, the constant being constrained not to change.

Explanation:

4 0

2 years ago

Explain why, when performing stoichiometric calculations, it is

pshichka [43]

Answer:

Stoichiometric Coefficients

The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element. Understanding this is essential to solving stoichiometric problems

Explanation:

3 0

2 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago

A teacher performs a flame test lab experiment for various compounds. During the lab, the teacher mislabels one of the solutions

nexus9112 [7]

Answer: By performing the flame test

Explanation:

The flame test can be performed by the teacher to visually identify the substance in the solution. The metals give a characteristic color on burning. The heat of the flame causes the excitation of the electrons present in the metal ions this leads to release of energy along with the emission of visible wavelength of light. This visible light can be observed by human eye. For example, the copper produces the blue color in flame, barium gives green flame, and sodium gives yellow flame.

4 0

2 years ago

234+34.1 add and round to 2 significant figures

marin [14]

Like this? 234+34.1= 268.1 then round. If it is less than 5 then you round down if it is more then you round up. Because it is less the final number would be 268.1=268

3 0

3 years ago