Answer:
9.17 atm
Explanation:
To find the new pressure of the gas, you need to use the following manipulated formula:
P₁V₁ / T₁ = P₂V₂ / T₂
In this formula,
P₁ = initial pressure (atm) P₂ = new pressure (atm)
V₁ = initial volume (L) V₂ = new volume (L)
T₁ = initial temperature (K) T₂ = new temperature (K)
Because you have been given values for all of the variables except for the new pressure, you can substitute them into the equation and simplify.
P₁ = 4.0 atm P₂ = ? atm
V₁ = 5.5 L V₂ = 2.0 L
T₁ = 300 K T₂ = 250 K
P₁V₁ / T₁ = P₂V₂ / T₂ <----- Given formula
(4.0 atm)(5.5 L) / (300 K) = P₂(2.0 L) / (250 K) <----- Insert variables
0.073333 = P₂(2.0 L) / (250 K) <----- Simplify left side
18.33333 = P₂(2.0 L) <----- Multiply both sides by 250
9.17 = P₂ <----- Divide both sides by 2.0
<h2>Answer:</h2>
<u>The right option is</u><u> (C) intermediate conductivity and a high melting point</u>
<h2>Explanation:</h2>
Metalloids usually look like metals but behave largely like nonmetals. Metalloids are shiny, brittle solids with intermediate good electrical conductivity. Their properties lie between metals and non metals. All metalloids exist as solids at room temperature and they have very high melting points. The physical properties of metalloids are more likely to be metallic, but their chemical properties tend to be non-metallic
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
I'm pretty sure it's called a volcanic winter, look up my answer to be correct though.
this way they can make sure that the experiment is correct.
