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3 years ago

How is carbon within the hydrosphere?

Chemistry

1 answer:

kap26 [50]3 years ago

8 0

Answer:

Carbon is found in the hydrosphere dissolved in ocean water and lakes. Carbon is used by many organisms to produce shells. Marine plants use cabon for photosynthesis. The organic matter that is produced becomes food in the aquatic ecosystem.

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An empty beaker weighs 39.09 g

Vinvika [58]

To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.

3 0

3 years ago

Heat transfer between two substances is affected by specific heat and the chemical composition of the substances. state of matte

lesya [120]

Answer:

the answer is D

Explanation:

7 0

3 years ago

Read 2 more answers

How is 8.2 x 10^4 - 6.3 x 10^3 written in scientific notation

ser-zykov [4K]

Answer:

= 7.57 × 104

(scientific notation)

= 7.57e4

(scientific e notation)

= 75.7 × 103

(engineering notation)

(thousand; prefix kilo- (k))

Explanation:

Just in case this is all of them

6 0

3 years ago

How many gallons of gasoline that is 5% ethanol must be added to 2,000 gallons of gasoline with no ethanol to get a mixture that

Amanda [17]

Answer:

$V_1= 3000 gal$

Explanation:

We have 3 solutions:

Solution 1 (with ethanol)
Solution 2 (no ethanol)
Final solution

$V_f=V_1 + V_2$

and for the ethanol:

$V_f*0.03=V_1*0.05 + V_2*0$

$V_f=V_1 \frac{5}{3}$

Combining:

$V_1 \frac{5}{3}=V_1 + V_2$

$V_1 \frac{2}{3}= V_2$

$V_1= \frac{3}{2} V_2$

If V2=2000 gal:

$V_1= \frac{3}{2} 2000 gal$

$V_1= 3000 gal$

8 0

3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago