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3 years ago

Discuss the transportation options now available in your community or region.

2 answers:

soldier1979 [14.2K]3 years ago

8 0

Answer: there are many transportation options. i had this once and all i did was list things like train car bike plane. then just list how they help

Explanation:

iogann1982 [59]3 years ago

3 0

Answer:

we have lots of bikes and i think that we could improve them to be portable scooters. we also have buses and i think that we should change them to subways to prevent all the pollution. Also i think cars should be less polluted in the future

Explanation: hope this is an okay answer.

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Describe the relationship between the light-dependent and the light-independent reactions.

Slav-nsk [51]

The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions.

Example:
NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle).

3 0

3 years ago

If you need some points and are good at science help!

bonufazy [111]

Answer:

region 2 and region 3

Explanation:

you can tell by the color of the land my friends^^

6 0

3 years ago

Read 2 more answers

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

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

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

A monomer is added to the end of a polymer chain in a __________ reaction.

aalyn [17]

Dehydration reaction

6 0

3 years ago

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A particular experiment requires 13.5 µl hydrochloric acid solution. if 250 ml hydrochloric acid solution is available, how many

ladessa [460]

The amount of HCl required for one experiment - 13.5 µl
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L
since one experiment uses up - 13.5 x 10⁻⁶ L
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times
the experiment can be carried out 18000 times

6 0

3 years ago

Read 2 more answers