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shutvik [7]
3 years ago
13

Plants can provide the materials that animals use in cellular respiration, and animals can provide some of the materials that pl

ants use for photosynthesis. The image below shows the relationship between photosynthesis and cellular respiration. According to the diagram, how does cellular respiration aid the process of photosynthesis?
It produces ATP.
It produces glucose.
It produces mitochondria.
It produces carbon dioxide
Chemistry
1 answer:
IrinaK [193]3 years ago
6 0

Answer:

it produce atp

Explanation:

.............

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Find the density.... Mass is 138g and Volume is 100 mL​
solniwko [45]

Answer:

1380 kilogram/cubic meter

p=\frac{m}{v}

=\frac{138g}{100mL}

=1.38

=1380

4 0
3 years ago
The Chemistry Cafe was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then,
maria [59]

In the question, we are told that there are;

  • A loaf containing 33 slices
  • A loaf containing 33 slices A package of cheese containing 15 slices

We also know that he is making a sandwich that has 2 pieces of both cheese and bread.

Hence;

Total number of bread and cheese = 33 + 15.

Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.

Therefore he can make = 33 + 15/4 = 12 sandwiches.

4 0
2 years ago
Read 2 more answers
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
Currently this lab is run qualitatively asking only for whether a solution is acidic, neutral, or basic. If quantitative informa
crimeas [40]

Answer:

The only PH range which is not covered by any of the given components of the universal indicator is 7.6-8.0

Hence the PH range 7.6-8 can't be described using universal indicator.

8 0
3 years ago
Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Selec
meriva

Answer:

Central atoms with four electron groups will be sp3 hybridized. True

Hybrid orbitals are delocalized over the entire molecule. False

The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True

Atoms with a single pi bond and an octet are sp2 hybridized. True

Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False

All resonance structures must be considered when assigning hybridization. False

Explanation:

When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.

Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.

According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.

When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.

Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.

The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.

8 0
3 years ago
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