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lys-0071 [83]
3 years ago
5

Despite the fact that the partial pressure difference is so much smaller for co2, why is there as much co2 exchanged between the

alveoli and blood as there is o2, ?
Chemistry
1 answer:
Snezhnost [94]3 years ago
7 0
This is due to the solubility of CO₂ in blood is more than the solubility of Oxygen in blood 
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Según la ecuación, HNO3 + H2S ----------- NO + S+ H2O el elemento que se oxida es
Serhud [2]

Answer:

HNO3 + H2S = H2O + NO + S - Chemical Equation Balancer. Balanced Chemical Equation. 2 HNO 3 + 3 H 2 S → 4 H 2 O + 2 NO + 3 S. Reaction Information. Nitric Acid + Hydrogen Sulfide = Water + Nitric Oxide (radical) + Sulfur . Reactants.

Explanation:

5 0
3 years ago
A. By referring to forces at work in the atomic nucleus, how can you
qwelly [4]

There is an unstable ratio of protons and neutrons. Because protons are positively charged but neutrons have no charge, an increase in the number of protons means there needs to be an increase in the number of neutrons to "bind" the nucleus together. This is because like charges repel, so the protons will repel each other, and if there aren't enough neutrons to act as "glue" to hold the nucleus together, the nucleus will break apart.

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2 years ago
In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is
V125BC [204]

Answer:

the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

Explanation:

Given that:

mass of diphenylacetylene (C_{14}H_{10}) = 0.5297 g

Molar Mass of diphenylacetylene (C_{14}H_{10}) = 178.21 g/mol

Then number of moles of diphenylacetylene (C_{14}H_{10})  = \frac{mass}{molar \ mass}

= \frac{0.5297  \ g }{178.24 \  g/mol}

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = Heat absorbed by H_2O + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene (C_{14}H_{10}) will be = \frac{20598.87 \ J}{0.002972 \ mol}

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

3 0
3 years ago
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Answer:

the answer is c kept in blue and with light

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RSB [31]
The answer is the second choice.
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