How many grams of calcium must be combined with phosphorus to form one mole of the compound Ca3P2?

29 Which term identifies a type of nuclear reaction?

AURORKA [14]

4) fission is a nuclear reaction where the nucleus of a reactant breaks apart

5 0

2 years ago

Read 2 more answers

How many hydrogen bonds can CH2O make to water

VladimirAG [237]

Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:

H2C=O---------H-OH 

But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. 

Formaldehyde in aqueous solution is in equilibrium with its hydrate. 

H2C=O + H2O <-----------------> H2C(OH)2

5 0

1 year ago

A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration

Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= $100$ ml

Initial moles of NO= $\frac{2.60}{10^2}$ moles

Initial moles of H2= $\frac{1.30}{10^2}$ moles

Concentration of NO at equilibrium= $0.161$ M

$Concentration(in M)=\frac{moles}{volume(in litre)}$

Moles of NO at equilibrium= $0.161(\frac{100}{1000})$

= $\frac{1.61}{10^2}$ moles

2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)

Initial :1.3*10^-2 2.6*10^-2 0 0 moles

Equilibrium:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles

∴ $\frac{2.60}{10^2}-x=\frac{1.61}{10^2}$

⇒ $x=\frac{0.99}{10^2}$

$Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)$

Equilibrium:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles

⇒ $Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2} (0.1)$

⇒ $Kc=19.4760$

3 0

2 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0

2 years ago

The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g

valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1 O ₂N N H ⁻ + H ⁺ ( f a s t e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ] / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

5 0

1 year ago