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Dmitriy789 [7]
3 years ago
15

Sublimation C

Chemistry
1 answer:
nadya68 [22]3 years ago
6 0

Answer:

A is Vaporisation

B is Sublimation

C is condensation

D is Deposition

E is Freezing

F is melting

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Acid Rain. Hope i helped you!
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Identify what type of reaction is below. Also determine what coefficients that would balance the chemical equation. ___ HCl + __
egoroff_w [7]

Answer:

Single Replacement; 2, 1, 1, 1

Explanation:

2HCl + Zn → H2 + ZnCl2

In a single replacement reaction, one specie is replaced by another specie in a chemical reaction.

If we look at the reaction above, zinc metal reacted with hydrochloric acid. The reaction proceeds in such a manner that zinc replaces hydrogen in hydrochloric acid. This is a single replacement reaction because only one chemical specie is displaced in the reaction.

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Photosynthesis allows plants to turn light energy into chemical energy by forming glucose from carbon dioxide and water: 6CO2(g)
dezoksy [38]

Answer:

305 g of CO₂

3.77 × 10⁵ kJ

Explanation:

Let's consider the global reaction for photosynthesis.

6 CO₂(g) + 6 H₂O(l) → C₆H₁₂O₆(g) + 6 O₂(g)  ΔHrxn = 2802.8 kJ

<em>A 1.70 lb sweet potato is approximately 73% water by mass. If the remaining mass is made up of carbohydrates derived from glucose (MW = 180.156 g/mol), how much carbon dioxide (MW = 44.01 g/mol) was needed to grow this sweet potato?</em>

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Let's consider the following relations:

  • The potato is 100%-73%=27% glucose by mass.
  • 1 lb = 453.59 g.
  • 6 moles of CO₂ produce 1 mole of glucose.
  • The molar mass of glucose is 180.156 g/mol.
  • The molar mass of carbon dioxide is 44.01 g/mol.

Then, for a 1.70 lb potato:

1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{6molCO_{2}}{1molGlucose} .\frac{44.01gCO_{2}}{1molCO_{2}} =305gCO_{2}

<em>How much light energy does it take to grow the 1.70 lb. sweet potato if the efficiency of photosynthesis is 0.86%?</em>

<em />

According to the enthalpy of the reaction, 2802.8 kJ are required to produce 1 mole of glucose. Then, for a 1.70 lb potato:

1.70lbPotato.\frac{27lbGlucose}{100lbPotato} .\frac{453.59gGlucose}{1lbGlucose} .\frac{1molGlucose}{180.156gGlucose} .\frac{2802.8kJ}{1molGlucose} .\frac{1}{0.86\% } =3.77\times 10^{5} kJ

7 0
3 years ago
What is the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution?
RoseWind [281]

Answer: 0.4M

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Given that,

Amount of moles of NaOH (n) = ?

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Since, n = mass in grams / molar mass

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Volume of NaOH solution (v) = 2.5 L

Concentration of NaOH solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

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c = 1 mole / 2.5 L

c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)

Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M

7 0
3 years ago
Solve for v in the formula d = v + at
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Answer: v=d−at

Explanation: I don't know if this will help

6 0
3 years ago
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