How many centimeters are there in 4.84 x 10 kilometers

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

4 0

2 years ago

n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are

kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

2 years ago

Under which conditions will sugar most likely dissolve fastest in a cup of water? A. a sugar cube with stirring at 5oC B. a suga

OverLord2011 [107]

I believe the answer is sugar crystals with stirring at 15 degrees Celsius.
Solubility is the maximum amount of a substance that will dissolve in a given amount of solvent at a specific temperature. There are two major factors that affect solubility are temperature and pressure. Temperature affects solubility of both solids and gases, but pressure only affects the solubility of gases. Increasing the surface area of solutes also increases the solubility.

5 0

2 years ago

Read 2 more answers

A gas occupies 72.1 at stp. At what temperature would the gas occupy 85.9 L at a pressure of 93.6 kPa?

Alika [10]

Answer:

328.1 K.

Explanation:

To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in.

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).

P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.

T₂ = P₂V₂T₁/P₁V₁ = (0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) = 328.1 K.

4 0

2 years ago