The following are bronsted- lowery acids. Determine what will form when each donates a proton. HI, H2O, NH4+, HNO3

1 answer:

6 0

1) Chemical reaction: HI(aq) → H⁺(aq) + I⁻(aq).
It gives an iodide anion.
2) Chemical reaction: H₂O → H⁺(aq) + OH⁻(aq).
It gives a hydroxide anion.
3) Chemical reaction: NH₄⁺(aq) → H⁺(aq) + NH₃(aq).
It gives ammonia.
4) Chemical reaction: HNO₃(aq) → H⁺(aq) + NO₃⁻(aq).
It gives nitrate anion.

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Elenna [48]

Answer:

Fe2(SO4)3 + 3BaCl2 → 2FeCl3 + 3BaSO4

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2 years ago

(((NEED ANSWER QUICK!!!))) Which is the stronger conjugate base, CN- or OCN-? Explain

Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN => Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

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2 years ago

A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere

kvv77 [185]

Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m = the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
$I_{g} = \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} = \frac{m}{12} \, kg-m^{2}$

According to the parallel axis theorem, the moment of inertia about the first axis is
$I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} = \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}$

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
$\frac{I_{g}}{I_{1}} = \frac{0.0833m}{0.1233m} =0.6756$

Answer: 0.676

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2 years ago

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How do you find the scientific notation of a Avogadro's number??

ruslelena [56]

Answer: ummmmmm hmmmmm

Explanation:

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2 years ago

The partial pressure of CO2 inside a bottle of soft drink is 4.0 atm at 25°C. The solubility of CO2 is 0.12 mol/L. When the bott

navik [9.2K]

We can solve this problem by using Henry's law.
Henry's law states that the amount of dissolved gas is proportional to its partial pressure.
$C=kP$
C is the solubility of a gas.
k is Henry's law constant.
P is the partial pressure of the gas.
We can calculate the constant from the first piece of information and then use Henry's law to calculate solubility in open drink.
0.12=4k
k=0.03
Now we can calculate the solubility in open drink.
$C_o=kP_o
$
$C_o=0.03 \cdot 3\cdot 10^{-4}=0.09\cdot 10^{-4} \frac{mol}{L}$
Now we need to convert it to g/L. One mol of CO2 is 44.01g.
The final answer is:
$C_o=0.09\cdot 10^{-4}\cdot 44,01=3.4\cdot 10^{-4} \frac{g}{L}$

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2 years ago