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3 years ago

What volume does .0685 mol of gas occupy at STP? 1mole =22.4

Chemistry

2 answers:

SashulF [63]3 years ago

6 0

Answer : The volume of gas at STP is, 1.53 liter

Solution : Given,

Moles of a gas = 0.0685 mole

As we know that at STP,

1 mole of gas contains 22.4 liter volume of gas

As, 1 mole of gas contains 22.4 liter volume of gas

So, 0.0685 mole of gas contains $0.0685\times 22.4=1.53L$ volume of gas

Therefore, the volume of gas at STP is, 1.53 liter

GuDViN [60]3 years ago

4 0

V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm

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Consider the following isotopic symbol: 137Ba2+

aliya0001 [1]

Answer:

a) 56 protons

b) 54 electrons

c) 81 neutrons

d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)

e)137Ba (with 56 protons and 81 neutrons)

f) atomic mass is 136.9 u ; the mass number is the sum of protons and neutrons and is 137

Explanation:

Step 1: Data given

137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.

For a neutral atom, the number of protons is equal to the number of electrons.

The different isotopes of an element have the same number of protons but a different number of neutrons.

137Ba2+ has 56 protons (this is the same as the atomic number)

137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)

137Ba2+ has 81 neutrons ( 137 - 56 = 81)

In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.

This isotope is 137Ba

Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

The atomic mass of 137Ba2+ is 136.9 u

The mass number is a count of the total number of protons and neutrons in an atom's nucleus.

The mass number of 137Ba2+ is 137

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3 years ago

What is an organism that survives on a hist organism and cuases harm to the host

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An organisam is part of your body plant life

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3 years ago

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A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

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4 years ago

BJECTIVE TYPE QUESTIONS

8090 [49]

Answer:

1.Physical change

2. Chemical change

3. Heat

4.photosynthesis

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3 years ago

How many grams of h2 are needed to produce 14.51 g of nh3?

gavmur [86]

Answer:

2.57 g of H₂

Solution:

The Balance Chemical Equation is as follow,

N₂ + 3 H₂ → 2 NH₃

According to Balance equation,

34.06 g (2 moles) NH₃ is produced by = 6.04 g (3 moles) of H₂

So,

14.51 g of NH₃ will be produced by = X g of H₂

Solving for X,

X = (14.51 g × 6.04 g) ÷ 34.06 g

X = 2.57 g of H₂

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3 years ago