The weight of the water lost is 0.693 g.
To calculate the weight of the water that is lost, we need to find the weight of the anhydrous salt. The anhydrous salt is the crucible with the added hydrate, minus the weight of the crucible. So, the weight of the anhydrous salt is 5.022 g - 3.715 g = 1.307 g. Since the weight of the hydrate is 2.000 g, the weight of the water that is lost is equal to the difference between the weight of the hydrate and the weight of the anhydrous salt, which is 2.000 g - 1.307 g = 0.693 g.
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Answer:
The mass per unit volume percentage is equal to 3.43%
Explanation:
given
mass=7 grams
volume = 2 * 102 ml = 204ml
The percentage mass per volume is given by
%(m/v) = w grams of solute * 100 / Volume of solution in ml
=
= 3.43%
The percent mass/volume (% m/v) for a solution containing 7.00 g of dextrose in 2.00×102 mL of solution is calculated as (7.00 g / 2.00×102 mL) * 100 = 3.47%, so the dextrose solution is 3.47% m/v.
The percent mass/volume (% m/v) is a way of expressing the concentration of a solute in a solution. It is calculated as the mass of the solute divided by the volume of the solution, multiplied by 100%. In this case, to calculate the % m/v for a solution that contains 7.00 g of dextrose in 2.00×102 mL of solution, you would use the following equation:
% m/v = (mass of solute / volume of solution) * 100
Substituting the given values into this equation, you get:
% m/v = (7.00 g / 2.00×102 mL) * 100 = 3.47%
Therefore, the dextrose solution is 3.47% m/v.
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Answer:
S²⁻(aq) + Cr²⁺(aq) ⇄ CrS(s)
Explanation:
The molecular equation includes all the species in the molecular form. Usually, it is useful to write this first to balance the equation. This is a double displacement reaction.
K₂S(aq) + Cr(NO₃)₂(aq) ⇄ 2 KNO₃(aq) + CrS(s)
The full ionic equation includes all ions and the species that no dot dissociate in water.
2 K⁺(aq) + S²⁻(aq) + Cr²⁺(aq) + 2 NO₃⁻(aq) ⇄ 2 K⁺(aq) + 2 NO₃⁻(aq) + CrS(s)
The net ionic equation includes only those ions that participate in the reaction and the species that do not dissociate in water.
S²⁻(aq) + Cr²⁺(aq) ⇄ CrS(s)
The net ionic equation for the precipitation reaction is: Cr+ + 3S → CrS(s)
The net ionic equation for the precipitation reaction between potassium sulfide and chromium(II) nitrate can be written as:
Cr + 3S → CrS(s)
In this reaction, the chromium(II) ions (Cr) react with the sulfide ions (S) to form chromium(II) sulfide (CrS) which precipitates as a solid.
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neutrons?
Answer:
Subtracting the number of protons from the atomic mass.
Answer:
Subtract Protons
Explanation:
Since the vast majority of atom's mass is found it's protons and neutrons, Subtracting the Number of Protons ( i.e. the atomic number ) from the atomic mass will give you the calculated number of neutrons in a atom,
Answer:
There are approximately 475 umol of chlorogenic acid in the sample.
Explanation:
The first step is indentifying the chlorogenic acid structure. As it can be seen in the figure attached, this molecule is a carboxylic acid containing just one carboxyl group. This means, that chlorogenic acid is a monoprotic acid and it is only able to donate one proton per molecule or one mol of protons per one mol of molecules.
The second step is to balance the titration equation. Considering that sodium hydroxide will generate one mol of hydroxyl ions per mol of salt, we can simplify the equation:
H⁺ + OH ⁻ → H₂O
Therefore, we now know that for each mol of NaOH consumed 1 mol of chlorogenic acid is titrated.
Thus, the last step is calculation the amount of NaOH consumed during the tritation. We can use the following equation:
In which C is the concentration, n the amount of moles and V the volume.
The result is that n = 475 umol.
Answer: 0.9851mol.
Explanation:
4.289 x 10^0 can be written as a regular number by moving the decimal point to the right or left based on the exponent value. Since the exponent is 0, the decimal point does not need to be moved. Therefore, the regular number form of 4.289 x 10^0 is simply 4.289.