<span>To solve this we assume that the gas inside the balloon is an ideal </span>gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant volume pressure and number of moles of the gas
the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:
T1/P1 = T2/P2
P2 = T2 x P1 / T1
P2 = 25 x 29.4 / 75
P2 = 9.8 kPa
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Answer:
a. Heterogeneous
b. Homogeneous
c. Homogeneous
d. Heterogeneous
e. Heterogeneous
Explanation:
A heterogeneous mixture is a mixture in which you can see multiple different ingredients in, for example vegetable soup, tea with ice and lemon slices, or fruit salad.
A homogeneous mixture is a mixture in which you can only see one thing, for example tea, seawater, or milk.
Explanation:
Initial Pressure = 24 lb in-2
Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)
Final Pressure = ?
Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)
No Change in Volume.
From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P1T1 = P2T2
P2 = P1T1 / T2
P2 = 24 * 268 / 308 = 20.88 lb in-2
There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
