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3 years ago

According to the law of superposition, where would you find objects with an older relative age?

2 answers:

7 0

Answer: if you think about a tree the older rings would be in the middle, so you’d find objects with older age below/under/inside

Greeley [361]3 years ago

6 0

Answer:

The law of superposition states that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock is older if it is farther down in the rock layers.

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Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?

lys-0071 [83]

Answer:

Explanation:

Given parameters:

pH = 3.50

Unknown:

concentration of [H₃0⁺] = ?

concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

pH = -log₁₀[H₃O⁺]

[H₃O⁺] = inverse log₁₀ (-pH) = $10^{-pH}$ = $10^{-3.5}$

[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

For the [OH⁻]:

we use : pOH = -log₁₀ [OH⁻]

Recall: pOH + pH = 14

pOH = 14 - pH = 14 - 3.5 = 10.5

Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

[OH⁻] = $10^{-pOH}$

[OH⁻] = $10^{-10.5}$ = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

8 0

4 years ago

Determine the mass in grams of 4.69 x 1021 atoms of barium. (The

Virty [35]

Answer:

1.07 g Ba

Explanation:

Hello there!

In this case, according to the definition of the Avogadro's number and the molar mass, it is possible to say that 6.022x10^{23} atoms of barium equal one mole, and at the same time, 1 mole equals 137.327 grams of this element; thus, it is possible to say that 6.022x10^{23} atoms of barium have a mass of 137.327 grams; therefore, it i possible for us to calculate the required mass in grams as shown below:

$4.69x10^{21}atoms*\frac{137.327gBa}{6.022x10^{23} atoms} \\\\=1.07gBa$

Best regards!

5 0

3 years ago

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (

Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^{x}[B]^{y}[C]^{z}$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}$

$\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}$

$3 = 3^{x} \\\\x =1$

Order w.r.t B : Use trials 2 and 5

$\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}$

$\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}$

$4 = 2^{y} \\\\y =2$

Order w.r.t C : Use trials 1 and 2

$\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}$

$1 = (0.5)^{z}$

z = 1

Therefore, order w.r.t C = 1

8 0

4 years ago

The question is below

rodikova [14]

The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0

3 years ago

Do all metals have more electrons than protons

Furkat [3]

Answer:

Explanation:

4 0

4 years ago