Answer:
The mineral forms gas bubbles when a sample is stored in water
<u>Answer:</u> The mass of solution that the chemistry student should use is 23.4 grams
<u>Explanation:</u>
We are given:
Available mass of isopropenylbenzene = 120. g
Amount of isopropenylbenzene needed by chemistry student = 10.00 g
42.7 % (w/w) solution of isopropenylbenzene.
This means that 42.7 grams of isopropenylbenzene is present in 100 grams of solution.
To calculate the mass of solution for given needed of isopropenylbenzene, we apply unitary method:
For 42.7 grams of isopropenylbenzene, the amount of solution needed is 100 grams
So, for 10.00 grams of isopropenylbenzene, the amount of solution needed will be = 
Hence, the mass of solution that the chemistry student should use is 23.4 grams
Answer:
The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case, you know:
- Q=?
- c= 4.186

- m= 12 g
- ΔT=Tfinal - Tinitial= 21 °C - 16°C= 5 °C
Replacing:
Q= 4.186
*12 g *5 °C
Solving:
Q=251.16 J
Since 1 J is equal to 0.2388 cal, then the following rule of three can be applied: if 1 J is equal to 0.2388 cal, then 251.16 J to how many cal are?

cal= 59.98 ≅ 60
<u><em>The heat required to raise the temperature of 12g of water from 16 C to 21 C is 60 cal.</em></u>
Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.
This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.
Next, let's determine how many moles of butane are burned. This is obtained by
5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.
Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.
We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).
V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.
Explanation: