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3 years ago

How long does it take for the entire pattern of moon phases to be completed?

Chemistry

2 answers:

o-na [289]3 years ago

8 0

B. It takes about a month

inn [45]3 years ago

8 0

The answer could be B

PS: Sorry if I’m wrong

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What is your guys ig?

makvit [3.9K]

Answer:

dont have one

Explanation:

7 0

3 years ago

Read 2 more answers

Carbon-14 has a half-life of 5,730 years. If scientists discover a mummified Cro-Magnon specimen, they will first want to determ

ryzh [129]

Answer:

the specimen has 11460 years old

Explanation:

if the live sample has as initial amount of Yo C14, the dead sample will have 0.25Yo of C14.

the rate of decay of radiactive matter over time is

Y(t) = Yo * e (( - t * Ln2 ) / T)

∴ Y(t) = 0.25Yo; T = 5730

⇒ 0.25Yo = Yo * e (( - t * Ln2 ) / 5730 )

⇒ 0.25 = e (( - t * Ln2 ) / 5730 )

⇒ Ln(0.25) = ( - t * Ln2 ) / 5730

⇒ - 7943.466 = - t * Ln2

⇒ 7943.466 / Ln2 = t

⇒ t = 11460 year

4 0

3 years ago

What are the concentrations of h3o+ and oh− in tomatoes that have a ph of 4.10?

IRINA_888 [86]

Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

pH : pH is defined as the negative logarithm of hydronium ion concentration.

Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

$4.10=-log[H_3O^+]$

$[H_3O^+]=antilog(-4.10)$

$[H_3O^+]=7.94\times 10^{-5}$

Now we have to calculate the pOH.

As we know, $pH+pOH=14$

$4.10+pOH=14$

$pOH=9.9$

Now we have to calculate the hydroxide ion concentration.

$pOH=-log[OH^-]$

$9.9=-log[OH^-]$

$[OH^-]=antilog(-9.9)$

$[OH^-]=1.258\times 10^{-10}$

Therefore, the concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

5 0

3 years ago

Find the boiling point temperature at 760 torr of an isomer of octane, c8h18, if its enthalpy of vaporization is 38,210 j mol-1

mixas84 [53]

According to Clausius-Clayperon equation,

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

$P_{1}$ is the vapor pressure at boiling point = 760 torr

$P_{2}$ is the vapor pressure at T_{2} =638.43 torr

Temperature $T_{2} = 110.0^{0}C + 273 = 383 K$

Δ $H_{vap} = 38210 J/mol$

Plugging in the values, we get

$ln(\frac{P_{1} }{P_{2} }) = \frac{delta H_{vap} }{R} (\frac{1}{T_{2} }-\frac{1}{T_{1} } )$

ln $\frac{760 torr}{638.43 torr} =\frac{38210 J/mol}{8.314 J/(mol.K)} (\frac{1}{383 K}-\frac{1}{T_{1} }$

$T_{1} = 389 K$

Therefore, the boiling point of octane = 389 K - 273 = $116^{0}C$

7 0

3 years ago

Lead all chlorate is mixed with hydrolylic acid. Each solution is 0.85 molar. Write balanced, molecular, ionic, and net equation

d1i1m1o1n [39]

Answer:

Here's what I get

Explanation:

Solubility rules

Salts containing halides are generally soluble. Important exceptions to this rule are halides of silver, mercury, and lead(II).
All acetates, chlorates, and perchlorates are soluble

So, PbCl₂ is insoluble, and Pb(ClO₃)₂ is soluble.

1. "Molecular" equation

$\rm Pb(ClO_{3})_{2}(aq) + 2HCl(aq) \longrightarrow \, PbCl_{2}(s) + 2HClO_{3}(aq)$

2. Ionic equation

Convert the soluble salts to their hydrated ions.

HCl and HClO₃ are strong acids. Convert them to their ions.

$\rm Pb^{2+}(aq) + 2ClO_{3}^{-}(aq)+ 2H^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + 2H^{+}(aq) + 2ClO_{3}^{-}(aq)$

3. Net ionic equation

Cancel all ions that appear on both sides of the reaction arrow (in boldface).

$\rm {Pb}^{2+}(aq) + \textbf{2ClO}_{3}^{-}(aq)+ \textbf{2H}^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + \textbf{2H}^{+}(aq) + \textbf{2ClO}_{3}^{-}(aq)$

The net ionic equation is

$\rm {Pb}^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s)$

4. Theoretical yield

We have the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

(i). Gather all the information in one place with molar masses above the formulas and masses below them.

M_r: 278.11

Pb(ClO₃)₂ + 2HCl ⟶ PbCl₂ + 2HClO₃

Volume/mL: 125 95

c/mol·L⁻¹: 0.85 0.85

(ii) Calculate the moles of each reactant

$\text{Moles of Pb(ClO$_{3}$)}_{2} = \text{0.125 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.1062 mol}\\\text{Moles of HCl} = \text{0.095 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.08075 mol}$

(iii) Identify the limiting reactant

Calculate the moles of PbCl₂ we can obtain from each reactant.

From Pb(ClO₃)₂:

The molar ratio of PbCl₂:Pb(ClO₃)₂ is 2:2

Moles of PbCl₂ = 0.1062 × 2/2 =0.1062 mol PbCl₂

From HCl :

The molar ratio of PbCl₂:HCl is 1 mol PbCl₂:2 mol HCl.

Moles of PbCl₂ = 0.08075 × 1/2 = 0.04038 mol PbCl₂

The limiting reactant is HCl because it gives the smaller amount of PbCl₂.

(iv) Calculate the theoretical yield of PbCl₂.

$\text{Theor. yield of PbCl}_{2} = \text{0.0438 mol} \times \dfrac{\text{278.11 g}}{\text{ 1 mol}} = \textbf{11.2 g}$

5. Calculate the actual yield of PbCl₂

$\text{Actual yield} = \text{11.2 g theor.} \times \dfrac{\text{ 68 g actual}}{\text{100 g theor,}} = \textbf{7.6 g}$

6. Calculate [ClO₃⁻]

Original concentration of Pb(ClO₃)₂ = 0.85 mol·L⁻¹

Original concentration of ClO₃ = 2 × 0.85 = 1.70 mol·L⁻¹

The solution was diluted by the addition of HCl.

Total volume = 125 + 95 =220 mL

c₁V₁ = c₂V₂

1.70 mol·L⁻¹ × 125 mL = c₂ × 220 mL

212.5 mol·L⁻¹ = 200 c₂

c₂ = (212.5 mL)/200 = 1.06 mol·L⁻¹

7. Calculate [Pb²⁺].

Moles of Pb²⁺ originally present = 0.1062 mol

Moles of Pb²⁺removed = 0.04038 mol

Moles of Pb²⁺ remaining = 0.0659 mol

c = 0.0659 mol/0.220 L = 0.299 mol·L⁻¹

8 0

4 years ago