Answer:
32g
Explanation:
potassium nitrate has solubility of about 67g per 100g of water at 40°C, which means that potassium nitrate solution will contain 67g of dissolved salt for every 100g of water.
since at this temperature, our solution contains 35g of potassium nitrate 100g of water. The solution will be unsaturated because of the less potassium nitrate.
to make saturated solution,
mass of potassuim nitrate = 67g - 35g
= 32g
which means dissolving another 32g of potassium nitrate in solution at 40
°C will make saturated solution.
I think the answer is research and development.
Answer: V₂ = 37.71mL
Explanation: To determine the new volume of Helium gas, use the Combined Gas Law, which states the following relationship among pressure, volume and temperature:
where index 1 relates to the initial state of the gas and index 2 to the final state of the gas.
Temperature is in Kelvin, so:
T = °C + 273
For this situation, standard pressure is 1 atm. Temperatures will be:
T₁ = 20 + 273 = 293 K
T₂ = 91 + 273 = 364 K
Solving:
37.71
The new volume of He gas is 37.71 mL.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
- m= 1.30 kg= 1,300 g (1 kg=1,000 g)
- ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.
Replacing:
Q= 64,121.2 J
<u><em>The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J</em></u>
Step 1
The reaction provided:
2 NaN3 (s) ⇨ 2 Na (s) + 3 N2 (g) (completed and balanced)
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Step 2
<em>Information provided:</em>
100.0 g NaN3 decomposes
-----
<em>Information needed:</em>
The molar masses of:
NaN3) 65.00 g/mol
N2) 28.00
(please, the periodic table is useful here)
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Step 3
Procedure
By stoichiometry,
2 NaN3 (s) ⇨ 2 Na (s) + 3 N2 (g)
2 x 65.00 g NaN3 ----------------- 3 x 28.00 g N2
100.0 g NaN3 ----------------- X
X = 100.0 g NaN3 x 3 x 28.00 g N2/2 x 65.00 g NaN3
X = 64.62 g
Answer: 64.62 g of N2 produced
