Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
2.2 moles H2O
Explanation:
, which rounds to about 2.2
Cohesion holds hydrogen bonds together to create surface tension on water. Since water is attracted to other molecules, adhesive forces pull the water toward other molecules.
Answer:
The correct answer is option B: "I, II".
Explanation:
The synthesis of 5-phosphoribosylamine from phosphoribosyl pyrophosphate (PRPP) is inhibited allosterically by ATP and GTP and activated by PRPP via feed forward activation. The enzyme that regulate this process is the enzyme amidophosphoribosyltransferase (AMPRT), which is stimulated by increased PRPP concentrations and is inhibited allosterically by ATP and GTP as well as IMP, AMP, and GMP.
<h2>Answer:</h2>
<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
