The true statement about basic solution at room temperature is that it has a greater concentration of hydroxide compared to hydronium ions.
Basic solutions have always pH greater than 7.
Basic solutions have bitter and caustic taste.
Basic solutions are not used as conductors in car batteries, acidic electrolytes are used in car batteries.
Answer:

Explanation:
The I₂ is the common substance in the two equations.
(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O
{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻
From Equation (1), the molar ratio of iodate to iodine is

From Equation (2), the molar ratio of iodine to thiosulfate is

Combining the two ratios, we get

A group of environmentalists were discussing the benefits and drawbacks associated with using fossil fuels. Which argument <span>best </span>fits the conversation?
Fossil fuels are cheaper than alternative forms of energy.Fossil fuel reserves will never be depleted.<span>Fossil fuels are easily renewed. </span><span>Fossil fuel use does not affect the environment.</span>
The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.
The reaction for the decomposition of water has been:

From the balanced equation, 2 moles of water decomposes to form 1 moles of hydrogen and 1 mole of oxygen.
The mass of oxygen produced has been 50 g. The moles of oxygen has been given by:

The moles of oxygen has been:

The moles of oxygen produced has been 1.5625 mol.
The moles of hydrogen decomposed has been given from the balanced chemical equation as:

The moles of hydrogen decomposes has been 3.125 mol.
The mass of hydrogen decomposed has been given by:

The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.
For more information about moles produced, refer to the link:
brainly.com/question/10606802
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)