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2 years ago

Which characteristic can be inherited?

Chemistry

1 answer:

FinnZ [79.3K]2 years ago

6 0

Answer:

long toes

Explanation:

the other traits are personality related, not a biological factor. they can be learned, but not inherited

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A chemist heats the block of copper as shown in the interactive, then places the metal sample in a cup of oil at 25.00 °C instea

Mazyrski [523]

When the oil is added to the heated copper, the energy in the system is

conserved.

The mass of the oil in the cup, is approximately 64.73 grams.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The mass of oil in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

m₂·c₂·(T₂ - T₃) = m₁·c₁·(T₃ - T₁)

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

The mass of the oil in the cup, m₁ ≈ 64.73 g

Learn more here:

brainly.com/question/21406849

Possible part of the question obtained from a similar question online, are;

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating = 65.17°C

6 0

2 years ago

Vanadium has an atomic mass of 50.9415 amu. It has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abu

grandymaker [24]

Answer:

Average atomic mass of the vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

Explanation:

6 0

3 years ago

How many amps are required to produce 75. 8 g of iron metal from a solution of aqueous iron(iii)chloride in 6. 75 hours?

Shalnov [3]

The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

I is the current passed

t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

Learn more about Law of Equivalence here, brainly.com/question/13104984

#SPJ4

7 0

1 year ago

An unknown metal displaces Ali from solution but does not displace sodium using the activity series determine the unknown metal

MAXImum [283]

Answer:

The reactivity of metal is determined by the reactivity series. ... The metal which easily displaced aluminium will lie above in the series but that same element cannot displace sodium, so it will lie below in the series. Hence, from the series, we conclude that the unknown metal could be calcium or magnesium.

Explanation:

Hope this helps! :)

8 0

2 years ago

(5.0 m3) ( 7.5 mmHg) = (P)(4.0m3)

anzhelika [568]

It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.

Given:

(5.0 m³)(7.5 mmHg) = (P)(4.0m³)

Multiply:

37.5 = 4.0P

Divide:

9.375 = P

P = 9.4 mmHg (remember sig figs)

<h3>Answer:</h3>

9.4 mmHg

7 0

3 years ago