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3 years ago

Major Battles of the

Chemistry

1 answer:

san4es73 [151]3 years ago

8 0

Answer: Lexington and Concord

Explanation: The British were trying to find weapons and could not find them.

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A naturally occurring element consists of three isotopes. The data for the isotopes are:

max2010maxim [7]

(46.972* 69.472% + 48.961*21.667% + 49.954*8.8610%)/100% =

= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u

7 0

3 years ago

Alpha particles are helium nuclei. electromagnetic waves. electrons. neutrons.

Firdavs [7]

Alpha particles are helium nucleus ! so answer is A !

it is made by ionization of helium, when both the electrons are out , then two unit of positive charge is one helium and thus called alpha particle !

4 0

3 years ago

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature

Stels [109]

<h3>Answer:</h3>

382.63 K

<h3>Explanation:</h3>

We are given;

Volume of Iodine as 71.4 mL
Mass of Iodine as 0.276 g
Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

We are going to use the ideal gas equation;

According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.

Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

= 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

= 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

3 0

3 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

3 years ago

Among the elements of the main group, the general trend in the first ionization energy moving across a period is not followed be

Radda [10]

The Ionization energy decreases because the full s orbital shields the electron entering the p orbital. This is known as "shielding", When each new electron experiences attraction from the nucleus and repulsion forces from the S orbitals, the net force on outer shell electrons is significantly smaller. Therefore, ionization energy decreases during these groups.

5 0

3 years ago

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