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MrMuchimi
3 years ago
7

The weight of the buggy was 40N on Mars. When the buggy landed on Mars it rested on an area of 0.025 m2. Calculate the pressure

exerted by the buggy on the surface of Mars
Chemistry
1 answer:
elena-s [515]3 years ago
4 0

Answer:

The pressure exerted by the buggy on the surface of Mars is 1600 pascals.

Explanation:

The pressure is determined by the definition of stress, which is the force exerted by the buggy on the martian surface divided by the contact area of the latter:

\sigma = \frac{F}{A}

Where:

\sigma - Stress, measured in pascals.

F - Force, measured in newtons.

A - Area, measured in square meters.

The force is the weight of the buggy (40 N) and A = 0.025\,m^{2}, the stress is now calculated:

\sigma = \frac{40\,N}{0.025\,m^{2}}

\sigma = 1600\,Pa

The pressure exerted by the buggy on the surface of Mars is 1600 pascals.

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Answer:

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Explanation:

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What is produced when 50mLs of carbon and 30mLs of oxygen react
dsp73

50ml of carbon dioxide.

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g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0
3 years ago
Science help me pleasee :)
kkurt [141]

Explanation:

i think right one is c.

it is because proton is different in each element.

6 0
3 years ago
1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to
solmaris [256]

Answer:

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

<em>2 moles of H₂ react per mole of O₂</em>

<em />

To find limiting reactant we need to convert the mass of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

10g H₂ * (1mol / 2.016g) = 4.96 moles

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

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For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
7 0
3 years ago
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