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notsponge [240]
2 years ago
5

In the 1980s, there was an international agreement to destroy all stockpiles of mustard gas, ClCH2CH2SCH2CH2Cl. When this substa

nce contacts the moisture in eyes, nasal passages, and skin, the ?OH groups of water replace the Cl atoms and create high local concentrations of hydrochloric acid, which cause severe blistering and tissue destruction. Complete and balance the equation for this reaction, and calculate ?H
Chemistry
1 answer:
Wittaler [7]2 years ago
5 0

Answer:

it messes with the H2O or water level in your skin

Explanation:

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AVprozaik [17]

Answer:

No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy

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2 years ago
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What is the predicted order of first ionization energies from highest to lowest for beryllium, calcium, magnesium, and strontium
Svetach [21]
We can predict the order of the elements given above according from the highest to lowest first ionization energies by using the trends in a periodic table. For elements in a family, the ionization energy decreases as it goes down. Therefore, the correct order would be Be, Mg, Ca, Sr.
6 0
2 years ago
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What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final
krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm)  of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}

P2 = 778.05 mm Hg = 1.02 atm

7 0
2 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
(9). A machine has a mechanical advantage of 0.6. What force should be applied to the machine to make it apply 600 N to an objec
Bogdan [553]
(9) Mechanical advantage = force by machine / force applied to machine
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F = 1000 N

(2) Efficiency = (output / input) x 100
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(5) Using the formula:
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h = 4.0 m

(6) Potential to electrical

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