In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

A triprotic acid is a type of Arrhenius acid that has the ability to donate three protons per molecule during dissociation in aqueous solutions. Thus, the chemical reaction, as outlined in the question, at the third equivalence point, can be expressed as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R denotes the counter ion of the triprotic acid. Consequently, the ratio of reacted acid to base at this point is 1:3.
The moles of NaOH are calculated as 0.106M*0.0352L = 0.003731 mole. Therefore, the amount of H3R is 0.003731mole/3=0.001244mole.
Subsequently, the molar mass of the acid can be determined: 0.307g/0.001244mole=247 g/mol.

<span>BaCl2 + Na2SO4 --> BaSO4 + 2NaCl In this reaction, 1.0 g of BaCl2 and 1.0 g of Na2SO4 are present. We need to identify the limiting reactant. "First, convert grams to moles" 1.0 g BaCl2 * (1 mol BaCl2 / 208.2 g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0 g Na2SO4 * (1 mol Na2SO4 / 142.04 g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2) = 1.5 mol Na2SO4 per mol BaCl2 "Using this ratio to compare with the balanced equation, BaCl2 + Na2SO4 --> BaSO4 + 2NaCl" The balanced equation indicates that 1 mol of BaCl2 reacts with 1 mol of Na2SO4. However, we found that 1.5 mol of Na2SO4 is available for each mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>

According to the periodic table:
the molar mass of barium is 137.2 grams
the molar mass of oxygen is 16 grams
the molar mass of hydrogen is 1 gram

The molar mass of Ba(OH)2 can be calculated as 137.2 + 2(16) + 2(1) = 171.2 grams.
The molar mass of 4H2O is computed as 4 [2(1) + 16] = 72 grams.

Consequently, the molar mass of Ba(OH)2·4H2O is 171.2 + 72 = 243.2 grams.

Therefore, a sample weighing 243.2 grams of <span>barium hydroxide tetrahydrate includes 72 grams of water, meaning that within 92.8 grams, the mass of water would be:
mass of water in 92.8 grams = (92.8 x 72) / 243.2 = 27.474 grams.

Thus, when heating a 92.8 gram sample of Ba(OH)2·4H2O (barium hydroxide tetrahydrate), 27.474 grams of water will be emitted.</span>

Answer:

Is it true that you shouldn't wash your hair, drink cold water, eat ice cream, or exercise during your period? Please explain your answer.

No, this is not accurate; doing any of these activities is perfectly fine. None of them affects us because they are not connected to our bodily systems. Also, I apologize for any language errors as I utilized Google Translate.

I hope this is helpful :)