Answer:
The percent by mass of water in this crystal is:
Explanation:
This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:
- 6,235 g = 100%
- 4.90 g = X
And the value of the variable X is found:
- X = (4.90 g * 100%) / 6,235 g
- X = approximately 78.6%.
The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:
- Water percentage = Total percentage - Percentage after heating.
- <u>Water percentage = 100% - 78.6% = 21.4%</u>
The arrangement of the solutions based on their absorption from highest frequency to lowest frequency :
b.
> c.
> a.NaCl
<h3>What is absorption frequency?</h3>
- The frequency of the molecular vibration that led to the absorption is the same as the absorption frequency of a basic IR absorption band.
- In a way, an emission spectrum is the opposite of an absorption spectrum.
- The discrepancies in the energy levels of each chemical element's orbitals correspond to absorption lines for each chemical element at various particular wavelengths.
- Therefore, it is possible to identify the constituents in a gas or liquid using its absorption spectrum.
- Absorption spectroscopy is most frequently used to measure infrared, atomic, visible, ultraviolet (UV), and x-ray waves.
Learn more about Absorption frequency here:
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Answer:
c
Explanation:
it is corundum now please follow me
One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
