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mario62 [17]
4 years ago
5

Which example best shows conservation of resources?

Chemistry
2 answers:
zaharov [31]4 years ago
6 0

Answer:

D.

Explanation:

I took this test as a freshman and this was the only question of 4 that I can remember

grin007 [14]4 years ago
4 0

。☆✼★ ━━━━━━━━━━━━━━  ☾  

Conservation means limiting waste of.

Therefore, the correct answer would be option D

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

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The chemical formula for strontium sulfite is SrSO3.<br><br><br> TRUE<br><br><br> FALSE
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Valency is the number of electrons lost or gained by an atom to attain an stable configuration. Valency is important when writing the formula of chemical compounds in chemistry. Strontium has a valency of  2 while sulfite ion (radicle) has a valency of 2. Therefore, the chemical formula of strontium sulfite is written as SrSO3.
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What energy is using the telephone?
kakasveta [241]

Answer:

electrical

Explanation:

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What is the molar mass of magnesium tartrate
Leya [2.2K]

Answer:

172.385 g/mol

Explanation:

Magnesium Tartrate is C4H4MgO6

C - 12.01 g/mol

H - 1.01 g/mol

Mg - 24.305 g/mol

O - 16.00 g/mol

12.01(4) + 1.01(4) + 24.305 + 16(6) = 172.385 g/mol

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if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met
Bumek [7]

ANS: density = 13.41 g/ml

Density (d) of a substance is the mass (m) occupied by it in a given volume (v).

Density = mass/volume

i.e. d = m/v

m = (d) v -----(1)

The given equation from the plot of weight vs volume is :

y = 13.41 x ----(2)

Based on equations (1) and (2) we can deduce that the density of the metal is 13.41 g/ml

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3 years ago
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A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca
Phoenix [80]

The specific heat capacity of the metal given the data from the question is 0.66 J/gºC

<h3>Data obtained from the question</h3>
  • Mass of metal (M) = 76 g
  • Temperature of metal (T) = 96 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 24.5 °C
  • Equilibrium temperature (Tₑ) = 31 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of the metal can be obtained as follow:

Heat loss = Heat gain

MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)

C × 4940 = 3263.52

Divide both side by 4940

C = 3263.52 / 4940

C = 0.66 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

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2 years ago
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