Answer:
The air that we breathe in enters the nose or mouth, flows through the throat (pharynx) and voice box (larynx) and enters the windpipe (trachea). The trachea divides into two hollow tubes called bronchi
Explanation:
Answer:
487.33 K.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant.
T is the temperature of the gas in K.
- If n is constant, and have two different values of (P, V and T):
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 5.4 atm, V₁ = 1.0 L, T₁ = 33°C + 273 = 306 K.
P₂ = 4.3 atm, V₂ = 2.0 L, T₂ =??? K.
<em>∴ T₂ = P₂V₂T₁/P₁V₁</em> = (4.3 atm)(2.0 L)(306 K)/(5.4 atm)(1.0 L) = <em>487.33 K.</em>
QUICK ANSWER
The name of the covalent compound N2O5 is dinitrogen pentoxide, more commonly known as nitrogen pentoxide. This covalent compound is part of a bigger group of compounds, nitrogen oxides, created purely from nitrogen and oxygen
Answer:
0.027 M HCl
Explanation:
The chemical equation of the neutralization is:
1 NaOH + 1 HCl -> 1 H2O + 1 NaCl
Because the ratio of NaOH and HCl is 1:1 you can use the M1V1=M2V2 formula.
(75 mL)(0.5 M NaOH) = (165 mL)(M HCl)
It requires 0.027 M HCl.
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
