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2 years ago

What kind of data does regional climate mainly collect and concentrate on?

Chemistry

2 answers:

hram777 [196]2 years ago

8 0

Answer:

Climate, atmosphere, and land

Explanation:

Some of the data collected include air chemistry, temperature, precipitation, cloud cover, and wind speed. Instruments carried on balloons and wind profiling radar provide observations from the surface to more than 10 miles high.

kupik [55]2 years ago

3 0

Climate researchers use every possible direct and indirect measurement to study the full history of Earth's climate, from the latest satellite observations to samples of prehistoric ice extracted from glaciers. When they focus on changes of the past 100-150 years, they use observations made by modern scientific instruments.

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Please help! Due today. Look at the image: Reuploaded

kirza4 [7]

Answer:

Explanation:

2+2=4-1=3 quick maths

8 0

3 years ago

Describe how a compound different from the elements that make it up?

Shkiper50 [21]

Answer: The difference between an element and a compound is that an element is a pure substance that is made of only one element, but a compound is 2 or more elements together

Explanation:

3 0

3 years ago

Read 2 more answers

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Musya8 [376]

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$ :

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

= $\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$ :

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{11 mol}{45 mol}\times 100=24.44\%$

7 0

4 years ago

Which pair of compounds has the same empirical formula? c3h8 and c3h c2h2 and c2h c4h10 and c6h c4h10 and c2h5.

BartSMP [9]

Answer:

The correct answer is the final pair: C4H10 and C2H5

Explanation:

Took the test and it was right. :)

8 0

3 years ago

The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5

Deffense [45]

Given: Half-life of 133Ba = t1/2 = 10.5 years.

The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.

Also, $k = \frac{2.303}{t} log \frac{Co}{Ct}$
where, Co = initial concentration = 1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................

Answer 1: For t = 5 years
 $0.066 = \frac{2.303}{5} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.

Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................

Answer 2: For t = 30 years
$0.066 = \frac{2.303}{30} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.

Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................

Answer 3: t = 180 years
$0.066 = \frac{2.303}{180} log \frac{Co}{Ct}$
Therefore, $log\frac{Co}{Ct}$ = 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.

Number of 133Ba atoms left after 180 years = 7.6367 X10^4.

8 0

4 years ago