5.4x10^3x1.2 x 10^7 in scientific notation

How can toxicant absorption be reduced after exposure to the skin? How can absorption be reduced for orally consumed chemicals?

zmey [24]

Answer:

The toxicant absorbtion can be reduced after exposure to the skin the surrounding clothing shoes or gloves should be removed or torn off than the part of the body which was exposured to the toxicant should be immediatly washed using clean running water for a while, with cold shower being the most recommended splashing method. In case of absorption for orally consumed chemicals should check on any remaining toxicant residue and be removed from the mouth. Vomiting should be induced to patients that are still conscious by providing them with liquids that can provoke vomiting. This will help in removing the toxicant in the intestinal and reduce their effect. Gastric lavage should then be done to induce diarrhea.

Explanation:

4 0

3 years ago

3.04 Quiz: Chemical Reactions 1 to 10 question

vaieri [72.5K]

Answer:

1. Chemical reactions

2.Substances

3. Properties

4. Precipitate

5. Light

6. Temperature

7. Color

8. Gas

Explanation:

During chemical reactions, new substances that possess new properties are formed. During the process of chemical reactions, evidences are found. Some of the evidences are formation of precipitate or light gel, production of gases and changes in color and temperature.

Chemical reactions start with the substances or compounds which are known as reactants. The reactants react to form new substances known as the products which possess new properties.

I only could help with 1 - 8 sorry.

5 0

3 years ago

5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(

professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

5 0

3 years ago

Read 2 more answers

Does salt affect how long an ice cube melts?

Mashutka [201]

Answer:

Ice cubes with salt melt faster. When you add salt, it dissolves in the water of the ice cube. The freezing temperature of salt water is 32 degrees Fahrenheit lower than that of fresh water.

5 0

3 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

guajiro [1.7K]

First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

= 1 mol * 18 mol/g

= 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

= 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

= 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

= 1.881 KJ +6.01 KJ + 4.514 KJ

= 12.405 KJ

5 0

3 years ago