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irakobra [83]
3 years ago
14

Mutations that neither benefit nor harm the organism have (blank) effect on the organism’s survival

Chemistry
1 answer:
Lisa [10]3 years ago
7 0

Answer:

neutral

Explanation:

neutral = no effect

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Which of the following is fact-based science rather than part of a belief system
makkiz [27]

Answer:

c. chemistry

Explanation:

it uses science and math which are based on facts.

6 0
3 years ago
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Order the terms according to the path followed by oxygen during cellular respiration.​
MArishka [77]

Answer:

Outside air

Nose

Lungs

Bloodstream

Cell

Explanation:

We breathe in oxygen from the outside air in through our nose and it travels to our lungs. Inside our lungs, we have Avioli's that diffuse oxygen into our bloodstream and the bloodstream helps the oxygen travel into our cells.

Hope this helps :)

8 0
3 years ago
In what ways do scientists not study the history of Earth's climate?
BARSIC [14]
(A) Studying ice cores
3 0
3 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
Estimate the number of moles of water in all the Earth's oceans. Assume water covers 75% of the Earth to an average depth of 3 k
o-na [289]

Answer:

there are approximately n ≈ 10²² moles

Explanation:

Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be

S= 4*π*R²

since the water covers 75% of the Earth's surface , the surface covered by water Sw is

Sw=0.75*S

the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is

V=Sw*D

the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is

m=ρ*V

the number of moles n of water ( molecular weight M= 18 g/mole = 1.8*10⁻² kg/mole ) for a mass m is

n = m/M

then

n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M

n=3*π*ρ*R² *D/M

replacing values

n=3*π*ρ*R² *D/M = 3*π*1000 kg/m³*(6.378*10⁶ m)² *3000 m /(1.8*10⁻² kg/mole) = 3*π*6.378*3/1.8 * 10²⁰ = 100.18 * 10²⁰ ≈ 10²² moles

n ≈ 10²² moles

8 0
3 years ago
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