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garik1379 [7]
2 years ago
10

Write a balanced chemical equation for the standard formation reaction of solid potassium dichromate k2cr2o7

Chemistry
1 answer:
almond37 [142]2 years ago
3 0
<span>A balanced chemical equation for the standard formation reaction of solid potassium dichromate is;
</span><span>K</span>₂<span>Cr</span>₂<span>O</span>₇→<span>K</span>⁺<span>(s) + KCr</span>₂<span>O</span>₇⁻<span>(s)</span>
Potassium dichromate with a chemical formula K₂Cr₂O₇ with molar mass 294.185 g/mol, is a common inorganic chemical reagent and its common use is as an oxidizing agent. It is odorless and its appearance is red-orange crystalline solid.
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If 4.59 g of potassium reacts with 3.6 g of sulfur according to the following reaction, how many grams of potassium sulfide can
EleoNora [17]
First convert the amount of grams you have of each substance to moles. Find your limiting reactant by calculating how many grams are needed to complete this reaction. If done correctly, you would see that we need .226 moles of Potassium to complete this reaction. However, we only have .118 moles of Potassium, so K must be our limiting reactant. Then use the moles of K to find out how many moles of K^2S are made. Then convert the amount of moles of K^2S to grams and you should get 10.3 g K^2S
8 0
2 years ago
Complete the equation for the dissociation of cdcl2(aq). omit water from the equation because it is understood to be present.
mixas84 [53]
Cadmium chloride is a highly soluble compound. The equation for its dissolution is:

CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
3 0
2 years ago
How much heat is lost when changing 65 g of water vapor (H2O) at 421 K to ice at 139 K?
poizon [28]

The heat change will be

Moles of water = mass / Molar mass = 65/ 18 = 3.61 mol

specific heat of ice =2.09J /g C

specific heat of water = 4.184 J/g C

Specific heat of vapour= 2.01 /g C

Heat of fusion = 3.33X10⁵ J /kg = 333 J /g

Heat of vaporization = 2.26 X10⁶J/kg = 2260J/g

Q1 = heat change when vapours get cooled to 373.15 K

Q2 = heat change when vapours get converted to liquid water

Q3 = heat change when liquid water cools to 273.15 K

Q4= heat change when liquid water freezes to ice

Q5= heat change when ice cools from 273.15K to 139 K

Q1= mass of water X specific heat of vapours X change in temperature

Q1 = 65 X 2.01 /g C X (421-373.15) = 6251.60 J = 6.252 kJ

Q2 = heat of vaporization X mass = 2260 X 65 = 146900 = 146.9 kJ

Q3 = mass X specific heat of water X change in temperature =

Q3 = 65 X 4.184 X (373.15-273.15) = 65 X 4.184 X 100 = 27196 J = 27.196kJ

Q4 = heat of fusion X mass =333X65 = 21645 J = 21.645 kJ

Q5 =  mass X specific heat of ice X change in temperature

Q5 = 65 X 2.09 X (273.15-139) = 18224.3 J = 18.224 kJ

Total energy = 6.252 +146.9+27.196+ 21.645+ 18.224 = 220.217

As this is energy released so it will be expressed in negative

-220.217

from the given options the correct answer will be -219.4 kJ

The answer is little different as the reference values of specific heats or enthalpy may vary.

3 0
2 years ago
1.
Diano4ka-milaya [45]

Answer:

The advantage of net ionic equations is that they show only those species that are directly involved in the reaction

Explanation:

3 0
2 years ago
Protein x has an absorptivity of 0.4 ml·mg-1 ·cm-1 at 280 nm. What is the absorbance at 280 nm of a 2.0 mg ·ml-1 solution of pro
Evgen [1.6K]

Absorbance measures the ability of the substance to absorb light at a specific wavelength.

Absorbance is also equal to the product of molar absorptivity, path length and molar concentration.

The mathematical expression is given as:

A= \epsilon l c       (1)

where, A = absorbance

\epsilon =  molar absorptivity

l = path length

c  = molar concentration.

The above formula is said to Beer's Law.

Absorptivity of protein x  = 0.4 mLmg^{-1}cm^{-1}

Path length = 1 cm

Molar concentration = 2.0 mg mL^{-1}

Put the values in formula (1)

Absorbance at 280 nm = 0.4 mL mg^{-1}cm^{-1}\times 1 cm \times 2.0 mg mL^{-1}

= 0.8

Thus, absorbance at 280 nm = 0.8

3 0
2 years ago
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