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2 years ago

4. What is the common name of the compound CaOCI,?

Chemistry

1 answer:

Paladinen [302]2 years ago

4 0

Answer:

Calcium Hypochlorite is an inorganic chemical compound with chemical formula CaOCl2. They are also called calcium oxychloride and the common name of CaOCl2 is bleching or chlorine powder.

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233.856 , sorry if i’m wrong

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2 years ago

How much work (in J) is involved in a chemical reaction if the volume decreases from 4.35 to 1.20 L against a constant pressure

Marianna [84]

Answer:

W = -262 J.

Explanation:

Hello there!

In this case, according to the given information, we can recall the definition of work in terms of constant pressure and variable volume as follows:

$W=P(V_2-V_1)$

So we plug in the given pressure and volumes to obtain:

$W=0.822atm(1.20L-4.35L)\\\\W=-2.60atm*L$

Now, we convert this number to J (Pa*m³) by using the shown below conversion factor:

$W=-2.60atm*L*\frac{101325Pa}{1atm} *\frac{1m^3}{1000L}\\\\W=-262J$

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2 years ago

In a chemical reaction, one of the chemical reactants is present in______ to make sure that the limiting reagent is completely c

Nata [24]

Answer: excess

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2 years ago

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur

dolphi86 [110]

Answer:

3.824 atm

Explanation:

From the ideal gas equation

P = mRT/MW × V

m is mass of testosterone = 12.9 g

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature of benzene solution = 298 K

MW is molecular weight of testosterone = 288.40 g/mol

V is volume of benzene solution = 286 ml = 286 cm^3

P = 12.9×82.057×298/288.4×286 = 3.824 atm

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2 years ago

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago

4. What is the common name of the compound CaOCI,?​

4. What is the common name of the compound CaOCI,?