Answer:
Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.
Explanation:
Atom refers to the smallest constituent unit of a chemical element. Molecules refer to a group of two or more atoms that are held together due to chemical bonds.
Answer:
0.486atm is the pressure of the cylinder
Explanation:
1 mole of Pb(NO₃)₂ descomposes in 4 moles of NO2 and 1 mole of O2. That is 1 mole descomposes in 5 moles of gas.
To find the pressure of the cylinder, we need to find moles of gas produced, and using general gas law we can determine the pressure of the gas:
<em>Moles Pb(NO₃)₂ and moles of gas:</em>
3.31g * (1mol / 331g) = 0.01 moles of Pb(NO₃)₂.
That means moles of gas produced is 0.05 moles.
<em>Pressure of the gas:</em>
Using PV = nRT
P = nRT/V
<em>Where P is pressure (Incognite)</em>
<em>V is volume (2.53L)</em>
<em>R is gas constant (0.082atmL/molK)</em>
<em>T is absolute temperature (300K)</em>
And n are moles of gas (0.05 moles)
P = 0.05mol*0.082atmL/molK*300K / 2.53L
P = 0.486atm is the pressure of the cylinder
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
Mass is the amount of matter in an object and does not change with location.
Explanation:
The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂
