Answer:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
Explanation:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m
<span>Energy emitted
Ee = hc/ 1736^-9m </span>
Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>
<span>Converting J to eV (1.60^-19 J/eV)
∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>
<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>
<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>
<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>
n² = -13.60 / -0.85 = 16
n = 4
<u>Answer:</u>
<u>Plasmas of great interest to scientists or manufacturers as</u>
- Plasma is electrically charged gases that contain considerable charged particles that can change the behavior of the substance.
<u>Current uses of plasmas:</u>
- First, it is used to make semiconductors for different types of electronic equipment
- Secondly, they're used in making transmitters for high-temperature films.
<u>Way scientists and engineers hope to use plasmas in the future:</u>
- The scientists are hoping to use plasma in the future to get rid of all hazardous wastes through a process called plasma gasification.
False, pepsin and hydrochloric acid
