Calculate the energy required to raise the temperature of 230. g of water from 20.0 ° C to 35.0 ° C.

Suppose you are a particle og watet in a lake. Describe what happens to you when a motorboat passes by.

Tamiku [17]

i will move and collapse with other h20 particles and splash around

3 0

4 years ago

. The electron affinity for an oxygen atom is negative. What statement is probably true regarding

kramer

The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>Regarding the second electron affinity for an oxygen, i. e., the electron affinity for O-, it is much larger and negative.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−

kolbaska11 [484]

Answer:

r = 3.61x $10^{-6}$ M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k. $[S2O2^{-8} ]^{x} x [I^{-} ]^{y}$

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

$\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}$

$\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y$

$0.67 = 0.67^x$

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

$\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}$

$\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y$

$0.33 = 0.5x 0.67^y$

$0.67 = 0.67^y$

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

$2.6x10^{-6} = kx0.018x0.036$ $kx6.48x10^{-4} = 2.6x10^{-6}$

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = $4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}$

r = 3.61x $10^{-6}$ M/s

7 0

3 years ago

Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/ato

nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by $N_v$ .

It is depends on

total no. of atomic number(N)
energy required for vacancy
Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
temperature (T).

$N_v=Ne^{-\frac{Q_v}{kT} }$

To find equilibrium number of of vacancies we have find N.

$N=\frac{N_A\ \rho}{A_{cu}}$

Here ρ= 8.45 g/cm³ =8.45 ×10⁶m³

$N_A$ = Avogadro Number = 6.023×10²³

$A_{Cu}$ = 63.5 g/mole

$N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}$

$=8.01\times 10^{28$ g/mole

Here $Q_v$ =0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

$N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})$

=2.34×10²⁴ vacancies/ mole

3 0

3 years ago

Identify the number of each type of atom in this formula 5C12H22O11<br><br>wdym

Lady bird [3.3K]

There are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in one molecule of C12H22O11.
In five molecules there are 60 carbon atoms, 110 hydrogen atoms, and 55 oxygen atoms.

6 0

3 years ago