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Rama09 [41]
3 years ago
13

How many molecules are there in 21.4 grams of Ca(OH)2?

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

The answer would be 0.288827452320528

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After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?
Bond [772]
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
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3 years ago
A rocket is designed to drop its first stage mid-flight. Due to a malfunction, this does not occur. What is a possible result of
artcher [175]

Answer:

The rocket is now too heavy to reach its destination.

3 0
4 years ago
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Fake answers are reported and your account will be banned
Leona [35]

Answer:

BRAINLIEST?

Explanation:

Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. My findings said that ammonia is a weak base, potassium hydroxide is a strong base, vinegar is a weak acid and ethyl alcohol is a weak acid.

Vinegar and ethyl alcohol are eliminated as they are acids. The question is on bases.... Potassium hydroxide is a strong base. So we are left with ammonia, being a weak base.

A is your answer

8 0
3 years ago
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Which is the correct name for p2o5? phosphorus dioxide phosphorus pentoxide diphosphorus pentoxide diphosphorus hexaoxide
Elenna [48]

Correct option:

The correct name for P_{2}O_{5}  is diphosphorus pentoxide.

Why P_{2}O_{5} is called diphosphorus pentoxide?

P_{2}O_{5} is commonly known as diphosphorus pentoxide.

Phosphorus pentoxide has an intriguing property in that P_{2}O_{5} is actually its empirical formula, whereas P_{4}O_{10} is its actual molecular formula.

However, the name of the chemical was obtained from its empirical formula rather than from its molecular formula. The official name for this substance is diphosphorus pentoxide.

Oxygen-containing binary compounds have "oxide" as their "last name." Phosphorus is the "first name."

We list each atom's numbers below:

The di- and Penta- prefixes are used to indicate the presence of two and five oxygen atoms, respectively, in the molecule.

Learn more about diphosphorus pentoxide here,

brainly.com/question/18237346

#SPJ4

6 0
2 years ago
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
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