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PolarNik [594]
2 years ago
12

Name 3 parts of an atom

Chemistry
1 answer:
yarga [219]2 years ago
3 0

Answer:

the three parts are protons, neutrons, and electrons

Explanation:

have a good day

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Identify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues.
AlexFokin [52]

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

7 0
3 years ago
Calculate the heat required to melt 7.35 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat =
grigory [225]

Answer:

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene = \frac{7.35 g}{78 g/mol}=0.09423 mol

Heat fusion of benzene,\Delta H_{fus} = 9.92 kJ/mol

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{fus}\times 0.09423 mol

=9.92 kJ/mol\times 0.09423 mol=0.9348 kJ=934.8 J

(1 kJ = 1000 J)

2) Heat vaporization of benzene,\Delta H_{vap} = 30.7 kJ/mol

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

Q=\Delta H_{Vap}\times 0.09423 mol

=30.7 kJ/mol\times 0.09423 mol=2.893 kJ=2,993 J

(1 kJ = 1000 J)

3 0
3 years ago
Is salt a covalently bonded<br>molecule
Andreyy89
No it’s ionic bond, because Na+ and cl-
6 0
3 years ago
If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3
yanalaym [24]
The correct answer would be 8.71 × 10<span>–10</span>
6 0
3 years ago
Chemical formula for solid lead II sulfide ore burns in oxygen gas the products are solid lead II oxide and sulfur dioxide gas
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2PbS(s) + 3O₂(g) = 2PbO(s) + 2SO₂(g)
8 0
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