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irakobra [83]
2 years ago
8

You want to determine the effect of a certain fertilizer on the growth of orchids grown in a greenhouse.

Chemistry
1 answer:
Aleks [24]2 years ago
3 0
Hypothesis - I predict that the orchids will grow best with a medium concentration of fertilizer
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What does the ideal gas law describe
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the law that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.

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Cleo has a vegetable garden and wants to increase the amount of nitrogen in the soil.
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The correct answer is the first option

Explanation: hope this helped

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1. Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O
vovikov84 [41]
1) Balanced equation

C3H8 + 5O2 -> 3 CO2 + 4 H2O

2) 0.700 L C3H8

Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio

1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2

0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8

x = 3.500 L O2

3) CO2 produced

1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>

x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2

4) Water vapor produced

1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>

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3 0
3 years ago
(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 12g Zinc Solid + 24g Silver Nitrate
VARVARA [1.3K]

Answer:

Zn + 2AgNO₃ → Zn(NO₃)₂ +2Ag

13.34g Zn(NO₃)₃ and 15.24g Ag are formed if reaction is 100% complete.

Explanation:

molar mass of Zn=65 g

molar mass of AgNO₃=170 g

molar mass of Zn(NO₃)₂ =189  g

molar mass of Silver = 108 g

Zn + 2  AgNO₃ → Zn(NO₃)₂ +2 Ag            eq(1)

1 mole Zn  reacts with 2 moles Silver nitrate to give 1 mole Zinc nitrate and 2 moles Silver

or

65 g Zn reacts with 2×170 g of AgNO₃  → 189 g Zn(NO₃)₂ and 2×108 g Ag    - eq(2)

First find limiting reagent of the reaction

65g Zn reacts with 2×170g of AgNO₃

12 g Zn reacts with (12÷65)×2×170 g AgNO₃

=62.7 g AgNO₃

For the reaction to go to 100% yield 12 g Zn will need 62.7 g AgNO₃

but amount of AgNO₃ is 24 g

So the reaction yields is limited by amount of AgNO₃.

AgNO₃ is the limiting reagent.

So calculate the yield of products with the amount of AgNO₃

by eq(2)

2× 170 gAgNO₃ gives 189 g Zn(NO₃)₃

=340 g AgNO₃ gives 189 g Zn(NO₃)₃

24 g AgNO₃ gives (24÷340) ×189 g Zn(NO₃)₃

=13.34g Zn(NO₃)₃

again by eq2

2×170 g AgNO₃ gives 2×108 g Ag

= 340 g AgNO₃ gives 216 g Ag

24 g AgNO₃ gives (24÷340)×216 g Ag

= 15.24g Ag

6 0
3 years ago
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