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3 years ago

Give three exaples of passive transport

Chemistry

2 answers:

OLga [1]3 years ago

5 0

Diffusion, facilitated diffusion, filtration, and osmosis

MA_775_DIABLO [31]3 years ago

3 0

Well the definition of passive transport is the movement of substances across a cell membrane without the use of energy by the cell

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C3H18 +<br> O2<br> +<br> CO2 + H2O<br> Balance equation

dimulka [17.4K]

Answer:

2C3H18 + 15O2 ---->6CO2 +18H2O

Explanation:

the number of reactant must be equal to the number of product.

7 0

3 years ago

Acid rain is produced when _______ mixes with the water of a raindrop.

goldfiish [28.3K]

The answer should be B - Nitrogen Dioxide.

5 0

3 years ago

5. Hydrogen & oxygen react chemically to form water. How much water

stepladder [879]

39.25 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂ + O₂ → 2 H₂O

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of H₂ = 14.8 / 2 = 7.4 moles

number of moles of O₂ = 34.8 / 32 = 1.09 moles

We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:

if 1 moles of O₂ produces 2 moles of H₂O

then 1.09 moles of O₂ produces X moles of H₂O

X = (1.09 × 2) / 1 = 2.18 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 2.18 × 18 = 39.25 g

Learn more about:

limiting reactant

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6 0

3 years ago

A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9

mamaluj [8]

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii) truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna: number of trucks needed/week= 134400 / 30000 = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

4 0

3 years ago

3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?

Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

$Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J$

Thus, the total energy turns out to be:

$Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J$

Best regards!

5 0

3 years ago