Answer:
Comet
Explanation:
Comets are the astronomical bodies of variable sizes which are dominant in space and are often considered as the giant dirty snowball. It is comprised of a mixture of substances such as ice, dust, and gases such as carbon dioxide (CO₂), ammonia (NH₃) and methane (CH₄). These comets usually have an average size of about 750 meters to about 20 kilometers in terms of diameter. These comets can sometimes enter into the earth's atmosphere due to the gravitational pull of the earth and can cause massive destruction on earth like the mass extinction event or the meteoric impacts. When they move in the earth's atmosphere, it forms a long tail.
The number of moles of gas, a balloon inflated to 7 L of volume at STP will be 0.31 moles.
<h3>What is the Molar volume ?</h3>
The molar volume of a gas is the volume of one mole of a gas at STP.
At STP, one mole (6.02×10²³ particles) of any gas occupies a volume of 22.4 L
Therefore,
If 1 mole of any gas occupies 22.4 L of volume at STP
Then, X mole of gas occupies 7 L of Volume at STP
Now,
Let's equate both the above conditions ;
X / 7 L = 1 / 22.4 L
X = 1/ 22.4 L x 7 L
X = 0.31 moles
Hence, the number of moles of gas, a balloon inflated to 7 L of volume at STP will be 0.31 moles.
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The density of an object is the mass of the object compared to its volume. The equation for density is: Density = mass/volume or D = m/v. Each substance has its own characteristic density because of the size, mass, and arrangement of its atoms or molecules.
Cannot describe their exact position.
Answer:
ΔS > 0 only for choice E: CH4(g) + H2O (g) → CO(g) + 3 H2(g)
Explanation:
Our strategy in this question is to use the trend in entropies :
S (solids) less than S (liquids) less than S (gases)
Also we have to look for the molar quanties involved of each state and their change to answer the question:
A. N2(g) + 3 H2(g) → 2 NH3(g)
Here we have 4 moles gases going to 2 moles of products, so the change in entropy is negative.
B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s)
The change in entropy is negative since we have 2 mol gases in the reactants and zero in the products.
C. CH3OH(l) → CH3OH(s)
A liquid has a higher entropy than a solid so ΔS is negative
D. False see A,B,C
E. The change in moles of gases is 4 - 2= 2, therefore ΔS is greater than O.