Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
Answer:
58g
Explanation:
In order to solve this problem, you must take a look at the solubility graph for potassium nitrate.
Now, the solubility graph shows you how much solute can be dissolved per 100g of water in order to make an unsaturated, a saturated, or a supersaturated solution.
You're looking to make a saturated potassium nitrate solution using
50g of water at 60∘C. Your starting point will be to determine how much potassium nitrate can be dissolved in 100g of water at that temperature in order to have a saturated solution.
As you can see, the curve itself represents saturation.
If you draw a vertical line that corresponds to 60∘C and extend it until it intersects the curve, then draw a horizontal line that connects to the vertical axis, you will find that potassium has a solubility of about
115g per 100g of water. Your answer is 58g of potassium nitrate
The pressure will continue to build up eventually causing a release of pressure or an explosion.
Answer:
237.5 K.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 5.2 atm).
V is the volume of the gas in L (V = 15.0 L).
n is the no. of moles of the gas in mol (n = 4.0 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = ??? K).
∴ T = PV/nR = (5.2 atm)(15.0 L)/(4.0 mol)(0.0821 L.atm/mol.K) = 237.5 K.
The RMS of O2 at 17 degrees is calculated as follows
RMs= ( 3RT/m)^1/2 where
R= ideal gas constant = 8.314
T= temperature= 17+273= 290 K
M= molar mass in KG = 32/1000= 0.032 Kg
Rms is therefore= sqrt (3x 8.314 x290/0.032 ) = sqrt( 226036.875
RMs=475.43