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Levart [38]
2 years ago
7

Lividity in a dead body is caused by capillaries decaying due to de-oxygenation.

Chemistry
2 answers:
sineoko [7]2 years ago
5 0

Livor mortis refers to the bluish-purple discoloration (lividity) under the skin of the lower body parts due to gravitation of blood after death. ... Dark pinpoint spots (Tardieu spots) are seen when capillaries rupture due to increased gravitational pressure (e.g. in the legs of a hanged person)

lions [1.4K]2 years ago
5 0

Explanation:  Lividity is unnatural color  or discoloration of the skin after death. It starts within 20-30 minutes of death, but is not visible by human eye until two hours. Amount of reduced haemoglobin determines the intensity of color. Capillaries bursts because of increased gravity.The body parts  which are in contact with any object or ground are not affected by this because capillaries are compressed.    

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Describe how to use Le Chatelier’s principle to predict the possible ways a chemical system can respond to changes.
Andre45 [30]
Le Chatelier's principle simply explains how equilibria change as you change the conditions of a reaction. If you have a reaction that is at equilibrium lets say (A + 2B <--> C + D) by removing C or D we can drive the reaction forward and products more products. I can provide a more in-depth description if needed.
4 0
2 years ago
Read 2 more answers
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak
Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0
2 years ago
Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0
2 years ago
A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?
kari74 [83]
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon, 
then, (415.2/89.2) ×100 =  465.47 g of the substance will be required to yield 34.6 moles of carbon.
8 0
2 years ago
Read 2 more answers
Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu
Nitella [24]

Answer:

\Delta \:H_{vap}=40383.88\ J/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)

Given:

P_1 = 13.95 torr

P_2 = 144.78 torr

T_1 = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K

T_1 = 298.15 K

T_2 = 75°C  = 348.15 K

So,

\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)

\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}

\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=40383.88\ J/mol

4 0
2 years ago
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