At equilibrium, the amount of product is equal to amount of reactant <br> True o False

What chemicals would be needed to create a mint that actually cleaned teeth while offering flavors and consumable chemicals? I'd

amid [387]

Answer:

ddddddddddddddddddddddddddddddddddddddddddddddddd

Explanation:

5 0

2 years ago

The graphics below shows how photosynthesis and cellular respiration are related.

lisov135 [29]

Options found from another source are:

a. oxygen. b. glucose. c. energy stored as ATP. d. carbon dioxide and water

Answer:

c energy stored as ATP

Explanation:

Cellular respiration converts glucose into energy in the form of ATP (c). The answer cannot be oxygen (a), because this is required for this process as a final electron acceptor. In terms of photosynthesis, oxygen is released as a by-product. The answer cannot be glucose (b) because that is our starting point for respiration, and what is synthesised during photosynthesis. The answer cannot be (d) as carbon dioxide and water are released by cellular respiration, and required by photosynthesis

8 0

3 years ago

Why is the temperature needed to freeze ocean water lower than the temperature needed to freeze the surface of a freshwater lake

shepuryov [24]

Ocean water freezes just like freshwater, but at Lower temperature. Fresh water freezes At 32°F but see water freezes at about 28.4°F because of the salt in it it can be melted down to use as drinking water

6 0

3 years ago

ASAP worth 15 points also

Taya2010 [7]

The person above me is correct I took a test on this so it’s the right answer

7 0

2 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

At equilibrium, the amount of product is equal to amount of reactant True o False