Answer:
the diaphragm and rib muscles contract
Explanation:
Inhalation is the set of movements that allows air to enter the lungs. In this process occurs the contraction of the diaphragm, causing its lowering. The intercostal muscles also contract causing the ribs to lift. This causes the chest to grow larger and the internal pressure of the lungs to become smaller than the external. With the reduction of intrapulmonary pressure, air ends up entering the airways and reaching the alveoli for gas exchange to occur.
Depends how much sugar are you talking about. If we are under the solubility concentrations of sugar no change or very little change can be observed in the volume of the solution. The sugar molecules will occupy the intermolecular spaces of the water molecules
218.4 grams of CaO is produced using 3.9 moles CaCO₃.
<h3>How we calculate weight of any substance from moles?</h3>
Moles of any substance will be defined as:
n = W / M
Given chemical reaction is:
CaCO₃ → CaO + CO₂
From the above equation it is clear that according to the concept of stoichiometry 1 mole of CaCO₃ is producing 1 mole of CaO. By using above formula, we calculate grams as follow:
W = n × M, where
n = no. of moles of CaO = 3.9 moles
M = molar mass of CaO = 56 g/mole
W = 3.9 × 56 = 218.4 g
Hence, 218.4 grams of Cao is produced.
<h3>How much grams do Cao's molecular weight equal?</h3>
Molecular weight of CaO. CaO has a molar mass of 56.0774 g/mol. Calcium Oxide is another name for this substance. Convert moles of CaO to grams or grams of CaO to moles. Calculation of the molecular weight: 40.078 + 15.9994 ›› Composition by percentage and element
<h3>How much calcium is required to create one mole of oxygen?</h3>
In order to create one mole of calcium oxide, the reaction between one mole of calcium (40.1 g) and half a mole of oxygen (16 g) is stoichiometric (56.1 g). This means that only 4.01 grams of calcium metal and 1.6 grams of oxygen can combine to generate 5.61 grams of calcium oxide.
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Answer:
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<em>m = </em><u><em>25</em></u><em> </em><em> </em><u><em>Kg</em></u>
Answer:
Wt Avg Atomic Wt = 28.0577 amu
Explanation:
Atomic Mass = ∑Wt Avg contributions of isotopes of an element
Si-28 => 92.21% => 0.9221 decimal fraction => Wt Avg Si-28 = 0.9221(27.9769) = 25.7695 amu
Si-29 => 4.69% => 0.0469 decimal fraction => Wt Avg Si-29 = 0.0469(28.9765) = 1.3590 amu
Si-30 => 3.10% => 0.0310 decimal fraction => Wt Avg Si-30 = 0.0310(29.9737) = 0.9292
Wt Avg Atomic Wt = ∑Wt Avg Contributions = (25.7695 + 1.3590 + 0.9292)amu = 28.0577 amu