Answer:
KHP + NaOH ⟶ KNaP + H₂O
Explanation:
The H in KHP is the acidic hydrogen. The OH⁻ in NaOH neutralizes it and converts it to H₂O.
The balanced equation for the reaction is
KHP + NaOH ⟶ KNaP + H₂O
Answer:
302 kj heat is released by lowering the temperature
Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
Radioactive isotope, also called radioisotope, radionuclide, or radioactive nuclide, any of several species of the same chemical element with different masses whose nuclei are unstable and dissipate excess energy by spontaneously emitting radiation in the form of alpha, beta, and gamma rays.
Answer:
3C3H8O2 and 16H2SO4
Explanation:
C3H8O2(aq) + K2Cr2O7(aq) → C3H4O4(aq) + Cr2(SO4)3(aq)
In acidic reaction you need to separate the half-reactions knows.
C3H8O2(aq) → C3H4O4(aq)
Cr2O7(2-) → Cr(+3) (aq)
Balance elements other than O and H. Add H2O to balance oxygen, balance hydrogen by adding protons (H+) and Balance the charge of each equation with electrons
C3H8O2(aq) + 2H2O(l) → C3H4O4(aq) + 8H(+) + 8e(-)
Cr2O7(2-) + 14H(+) + 6e(-) → 2Cr(+3) + 7H2O (aq)
Scale the reactions so that the electrons are equal. If not look for a common multiple. In this case is 24, 8*3=6*4. So for each equation multiply for the correct multiple. Then
3C3H8O2(aq) + 6H2O(l) → 3C3H4O4(aq) + 24H(+) + 24e(-)
4Cr2O7(2-) + 56H(+) + 24e(-) → 8Cr(+3) + 28H2O (aq)
Add the reactions and cancel out common terms and group the ions in compounds if u can.
3C3H8O2(aq) + 32H(+) + 4Cr2O7(2-) → 3C3H4O4(aq) + 4Cr2(SO4)3(aq) + 22H2O (aq)
We are in acidic medium so the H(+) must be from H2SO4. So 32H(+) is 16H2SO4 in reactives. Add SO4(2-) in products and balance
16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + 4SO4(2-)
Then
16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + + 2K2SO4 (aq)
