Answer:
2.67 × 10⁻²
Explanation:
Equation for the reaction is expressed as:
CaCrO₄(s) ⇄ Ca₂⁺(aq) + CrO₂⁻⁴(aq)
Given that:
Kc=7.1×10⁻⁴
Kc= ![[Ca^{2+}][CrO^{2-}_4]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5BCrO%5E%7B2-%7D_4%5D)
Kc= [x][x]
Kc= [x²]
7.1×10⁻⁴ = [x²]
x = 
x = 0.0267
x = 
At 12 mph, how long does it take to go 13.1 miles?
We know that distance = rate * time
So we know that 13.1 = 12 * time
We can now see that time = 13.1/12 = 1.092 (hours)
That's approximately 1 hour and .092*60=5.52 minutes.
The molecular formula of methylpropan-1-ol is C4H10O, so the complete combustion equation is: C4H10O + 6O2 --> 4CO2 + 5H2O. This mean to completely combust 1.0mol of methylpropan-1-ol, 6 mol of O2 is required. Molar mass of O2 is 32 g/mol, so 32g/mol x 6mol = 192 g of O2 is required. At room temperature and pressure, the density of O2 is 1.3315 g/L (this can be obtained by density of gas = P/RT). So the volume of O2 = mass/density = 192g/1.3315(g/L) = 144 L = 144 dm3. The answer is B.
Increase the force they are using to move the box by adding more helpers
Explanation:
If you want to increase the speed that the box is moving, increasing the friction would make it more difficult, decreasing the force they are using would just make it slower and increasing the angle of the ramp would make it harder. Adding more people would make it easier. “The more the merrier”
