A = k (b/c)'k' is some constant number.
Angle and its complement have a sum of 90 degrees: A+C = 90 Angle plus five times its complement is 298 degrees: A+5C = 298 Subtract first equation from the second: 4C = 208 C = 52 So, the complement is 52 degrees and the angle is 38 degrees
B, C, D, E, H, I, K, O, X, (B, C and K not in all fonts) c, l, o, x
use this strategy: integral of (b^x) dx = (b^x)/ln(b) + K [K is integration constant, b is not a variable]rewrite (1/c)^(1-x) = ((1/c)^1)*((1/c)^(-x)) = (1/c)*(c^x). (1/c) is a constant, so bring outside the integral, then let b = c in the formula above, and you have (1/c)*(c^x)/ln(c) + K
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Oxygen is a gas at 298 K.
add 273 to get from C to K 25 + 273 = 298 Kelvins
298 degrees Fahrenheit = 420.93 K.
Chlorine is a gas at room temperature and pressure. Its melting point is −101.5 °C while its boiling point is −34.04 °C. So at 25 °C, it would be in the gaseous state.
An exothermic reaction is one where heat is released to the surroundings. An example of an exothermic reaction equation at 298 K is: 2H2(g) + O2(g) -> 2H2O(l) + heat
To convert Fahrenheit to Kelvin, you can use the formula K = (F - 32) x 5/9 + 273.15. Plugging in 298 degrees Fahrenheit gives you K = (298 - 32) x 5/9 + 273.15, which equals approximately 425.37 Kelvin.
If we assume "room temperature" to be 25 °C (77 °F) then that would equate to 298.15 K. Different areas assume different standard "room temperature" - but it usually falls between 20 °C (293.15 K) and 30 °C (303.15 K)
Use this equation to convert Kelvin to degrees Celsius/Centigrade: [°C] = [K] - 273.15
Using the ideal gas law, we can calculate the new volume of the balloon. Given that the initial volume is 0.75 L, the initial temperature is 25°C (298 K), and the final temperature is -100°C (173 K), we can use the equation (V1/T1) = (V2/T2) to find the new volume. Plugging in the values, we get V2 = (0.75 L * 173 K) / 298 K ≈ 0.44 L.
It is approximately 25°C. In Fahrenheit it is approximately 78°F It is also approximately 298 K
The value of Kw (ion product of water) at 298 K is approximately 1.00 x 10^-14.
To solve this problem, we can use the combined gas law formula: (P1 × V1) / (T1) = (P2 × V2) / (T2) Plugging in the values, we get: (1.20 ATM × 150.0 mL) / (298 K) = (1.50 ATM × V2) / (293 K) Solving for V2, we get V2 ≈ 120.0 mL. Therefore, the gas sample would have a volume of approximately 120.0 mL at 1.50 ATM and 20.0°C.