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mart [117]
3 years ago
11

an object is placed at 0 on a number line.it moves 3 units to the right,then 4units to the left,and then 6 units to the right th

e displacement of the object is?
Chemistry
2 answers:
Soloha48 [4]3 years ago
8 0

The correct answer is: +5

Explanation:

An object is placed at 0; it means:

Initial position of the object = 0.

Now it moves to 3 units to right, so keeping the standard cartesian coordinate system in mind (in right right x-axis is positive and left x-axis is right), the new position of the object will be +3.

Object now moves 4 units to the left, it means +3 - 4 = -1; object is at the position -1.

Object then moves 6 units to the right, therefore,

Final position of the object = -1 + 6 = +5.

Displacement = Final position - Initial position

Displacement = +5 - 0 = The correct answer is: +5

Evgen [1.6K]3 years ago
4 0
It should probably be on a positive 5
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mylen [45]
In the reaction Sn(s) + 2H+(aq) → Sn2+ (aq) + H2(g)
from this reaction, we get that Sn loses from 0 to 2 electrons so it's oxidized So it is the reducing agent.
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6 0
3 years ago
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Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
4 years ago
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GREYUIT [131]
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8 0
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Answer:

D. The equipment needed to accommodate the high temperature and pressure will be expensive to produce.​

Explanation:

Hello!

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In such a way, the high pressure favors the formation of ammonia whereas the high temperature the formation of hydrogen and nitrogen and therefore, option A is ruled out. Since the high pressure shifts the reaction rightwards and the high temperature leftwards, we would not be able to know whether the reaction has ended or not because it will be a "go and come back" process, that is why B is also discarded. Now, since hydrogen and nitrogen would be the "wastes", we discard C because they are not toxic. That is why the most accurate answer would be D. because it is actually true that such equipment is quite expensive.

Best regards!

6 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION!!
Ksenya-84 [330]
2 I think not 100% tho
5 0
3 years ago
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