What forces are responsible for anticlines and synclines

Calculate the wavelength of light with a frequency of 1.89 x 1018 Hz.

Charra [1.4K]

Which sentence from "Black Hole Beginnings" contains a cause-and-effect relationship?
More matter mounts up on the outside, increasing the pressure on the inside.
Instead it is a dramatic melding, or blending, of atoms, called nuclear fusion.
Each atom has its own tiny center, or nucleus.
Surrounding each nucleus is an electrical force field..

6 0

4 years ago

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each

Phoenix [80]

Answer:

Balanced reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Reduced: O₂

Oxidized : C₃H₈

Explanation:

For the given reaction:

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

In the reactants, there are 3 C, and in the products only one, so we multiply CO₂ by 3:

C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)

In the reactants, there are 8 H, and in the products, there are 2 H, so we multiply H₂O by 4:

C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)

In the reactants are 2 O, and in the products 10 O, so we multiply O₂ by 5:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The reaction is balanced.

Let's identify the oxidation number (Nox) of the elements in each compound.

C₃H₈:

H has fix Nox equal to +1, and the molecule is neutral, so, calling x the Nox of C:

3x + 8 = 0

3x = -8

x = -8/3

O₂:

Because is a pure compound, the Nox of O is 0.

CO₂:

The Nox of O is fix equal to -2, and the molecule is neutral, so, calling x the Nox of C:

2x -4 = 0

2x = 4

x = +2

H₂O:

The Nox of H and O are fixed, respectively, +1 and -2.

So, the carbon in C₃H₈ is oxidized because its Nox is increasing, and oxygen in O₂ is reduced because its Nox is decreasing.

6 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago

If a radioactive decay produced 1.26 × 1012 J of energy, how much mass was lost in the reaction?

Ket [755]

The mass lost in the process is 1.4 * 10^-5Kg.

<h3>What is the Einstein equation?</h3>

The Einstein equation is able to relate the mass lost to the energy that is evolved in a nuclear process.

Given that;

E = Δmc^2

Δm = E/c^2

Δm = 1.26 × 10^12 J/( 3 * 10^8)^2

Δm = 1.4 * 10^-5Kg

Learn more about Einstein equation:brainly.com/question/10809666

#SPJ1

8 0

2 years ago

The camel stores the fat tristearin (C57H110O6) in its hump. As well as being a source of energy, the fat is also a source of wa

Shalnov [3]

Answer:

8.1 × 10² g

Explanation:

Step 1: Write the balanced equation

2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O

Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g

We will use the conversion factor 1 lb = 453.592 g.

1.6 lb × 453.592 g/1 lb = 7.3 × 10² g

Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆

The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.

7.3 × 10² g × 1 mol/890.83 g = 0.82 mol

Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆

The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol

Step 5: Calculate the mass corresponding to 45 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

45 mol × 18.02 g/mol = 8.1 × 10² g

4 0

3 years ago