Avogardo's Law
Formula for the nth term of general geometric sequence tn = t1 x r(n - 1) For n = 2, we have: t2 = t1 x r(2 - 1) t2 = t1r substitute 11.304 for t2, and 2.512 for t1 into the formula; 11.304 = 2.512r r = 4.5 Check:
The given sequence is an arithmetic sequence with a common difference of -4. To find the nth term formula, we first determine the first term, which is 100. The nth term formula for an arithmetic sequence is given by: a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the term number, and d is the common difference. Therefore, the nth term formula for this sequence is a_n = 100 - 4(n-1) or a_n = 104 - 4n.
Flipping a coin: two possible outcomes, H or T. Rolling a die: six possible outcomes, 1, 2, 3, 4, 5, or 6. Flipping a coin and rolling a die: 12 possible outcomes. So the sample space has 12 outcomes such as, {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }
To find the sum of 25 terms of these arithmetic sequence you can use the formula:Sn = (n/2)(a1 + an), where n is the number of terms in the sequence, a1 is the first term, and an is the last term of the sequence. In our case n = 25, so we need to compute a1 and a25.Since an = 5t - 3, thena1 = 5(1) - 3 = 5 - 3 = 2a25 = 5(25) - 3 = 125 - 3 = 122By substituting the values we know into the formula we have:S25 = (25/2)(2 + 122) = (25/2)(124) = 25 x 62 = 1,550Or you can use the formula:Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference.In order to find d, we need to find at least the value of 2 terms and subtract them.a1 = 2a2 = 5(2) - 3 = 10 - 3 = 7So d = 7 - 2 = 5By substituting the values we know into the formula we have:S25 = (25/2)[2(2) + (25 - 1)5]S25 = (25/2)(4+ 120) = (25/2)(124 = 25 x 62 = 1,550Thus, the sum of 25 terms of the given arithmetic sequence is 1,550.
Avogardo's Law
The general representation of the combined gas law is P1V1/T1 = P2V2/T2
The Combined Gas Law relates pressure (P), volume (V) and temperature (T). The appropriate SI units are P in atm, V in liters, and T in degrees Kelvin. The Combined Gas Law equation is (P1*V1)/T1 = (P2V2)/T2. Isolating for V2 the equation then becomes (P1V1T2)/(T1P2) = V2
Because it is so high up the air. Pressure up there is high and temperature therefore low if you use P1V1/T1= P2V2/T2
Using the combined gas law (P1V1/T1 = P2V2/T2), we can calculate the new volume of the oxygen gas sample at 600K. Given P1V1/T1 = P2V2/T2, we have P1 = P2 (pressure is constant), V1 = 150 mL, T1 = 300K, and T2 = 600K. Plugging in these values, we get V2 = (P1 * V1 * T2) / (T1) = (1 * 150 * 600) / (300) = 300 mL. So, the new volume of the oxygen gas sample at 600K would be 300 milliliters.
From Boyle ideal gas law P1V1/T1 = P2V2/T2 so volume is reduced by a factor of 4
To find the original volume of gas, you can use the combined gas law equation: (P1V1)/T1 = (P2V2)/T2. Plug in the given values to solve for the original volume, where P1 = 135 kPa, V1 = 575 L, T1 = 295 K, P2 = 105 kPa, and T2 = 270 K. Solving for V1 will give you the original volume of the gas.
Using the combined gas law, we can find the pressure by setting up the equation: P1V1/T1 = P2V2/T2. At STP, T1 = T2, so the equation simplifies to P1V1 = P2V2. Plugging in the values, we get P1 = (2.875 L * 1 atm) / 6.25 L = 1.38 atm.
P1V1/T1 = P2V2/T2Assuming only temperature and volume are changing and pressure will be kept constant:V1/T1 = V2/T2Only Kelvin can be usedV1/273 = V2/523Assume the volume at 0 ºC is 1 unit thenV2 = 1.92 units
If it is an Ideal Gas, then you can use: P1V1/T1 = P2V2/T2, and since volume is constant, you have P1/T1 = P2/T2, where P is pressure and T is absolute temperature, and the subscripts refer to the 1st state and the 2nd state of the gas.So: (7.00 atm)/(379 K) = P2/(425 K), solve for P2 =(425 K)(7.00 atm)/(379 K)= 7.85 atm
Using the combined gas law, we can calculate the new pressure. The initial conditions are V1 = 900 mL, P1 = 2.50 ATM, and T1 = 298 K. The final conditions are V2 = 450 mL, T2 = 336 K. Using P1V1/T1 = P2V2/T2, we can solve for the new pressure, which is around 3.33 ATM.
The volume is 0,446 L.