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3 years ago

Help asap i need help

Chemistry

1 answer:

zalisa [80]3 years ago

3 0

Answer:

from producers to consumers

Explanation:

Energy is transferred between organisms in food webs from producers to consumers. The energy is used by organisms to carry out complex tasks. The vast majority of energy that exists in food webs originates from the sun and is converted (transformed) into chemical energy by the process of photosynthesis in plants

Short answer, don't know but i'm sure the answer is up here

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What is the mass of an object that has a density of 2.3 g/mL And a volume of 343ML

Bogdan [553]

Answer:

788.9g

Explanation:

density=m/v

2.3g/ml=m/343ml

=343×2.3

=788.9g

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3 years ago

1) How much energy would be required to melt 15.0 g of ice at 0°C?

DerKrebs [107]

Answer:

5010J

Explanation:

The following data were obtained from the question:

Mass (m) = 15g

Heat of fusion (ΔHf) = 334J/g

Heat required (Q) =..?

The heat energy required to melt the ice can be obtained as follow:

Q = m·ΔHf

Q = 15 x 334

Q = 5010J

Therefore, the heat energy required to melt the ice is 5010J.

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3 years ago

9. ______________ Consists of the Sun and everything that orbits the Sun

Rudiy27

Answer:

9.solar system

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3 years ago

What is the molar mass of sodium carbonate (Na2CO3)?

madreJ [45]

Answer:

105.9888 g/mol

Explanation:

The molar mass of sodium carbonate is 105.9888 g/mol (grams per mole)

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4 years ago

Read 2 more answers

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

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3 years ago