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3 years ago

On a victim, a circular entrance wound with black edges is found. Which most likely caused this wound?

Chemistry

2 answers:

castortr0y [4]3 years ago

4 0

The answer is D. my friend :)

max2010maxim [7]3 years ago

4 0

a pistol fired at close range with leave the black mark on the skin the correct answer is d

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A 10.00 mL sample of vinegar (an aqueous solution of acetic acid) is titrated with 0.5062 M NaOH(aq) and 16.58 mL is required to

sammy [17]

Answer:

5.01%

Explanation:

Density of vinegar = mass/volume

Mass of 10.00 mL = density x volume

= 1.006 x 10 = 10.06 g

From the equation of reaction:

$CH_3COOH(aq)+NaOH(aq)-->CH_3COONa(aq)+H_2O(l)$

1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.

mole of NaOH = molarity x volume

= 0.5062 x 0.01658

= 0.008392796‬ mole

0.008392796‬ mole of NaOH will therefore require 0.008392796‬ mole of CH3COOH.

mass of CH3COOH = mole x molar mass

= 0.008392796‬ x 60.052

= 0.504 g

Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%

= 5.01%

The percent by mass of acetic acid in the vinegar is 5.01%

3 0

2 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

How many molecules of glucose are produced by each cycle of the light reaction?

disa [49]

Zero (0) molecules of glucose are produced.

4 0

3 years ago

Which is not a type of mass movement

lana66690 [7]

Answer:

options?

Explanation:

3 0

2 years ago

Read 2 more answers

Many times, during a chemical reaction, energy is given off. this type of chemical reaction can be termed

WITCHER [35]

Physical can be seen

3 0

3 years ago