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3 years ago

4.564000 x 10^-4 express how many significant figures

Chemistry

1 answer:

Vaselesa [24]3 years ago

5 0

Answer: 7

Explanation:

Before a number but after a decimal. The zeros at the end would usually mean that it doesn't count but since the numbers are before the zeros and after a decimal it's 7 sig figs

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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is

lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄

64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

0.6 mol CH₄ * $\frac{2molH_2O}{1molCH_4}$ = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

1.2 mol * 18 g/mol = 21.6 g

4 0

2 years ago

For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a

Volgvan

<span>For this reaction, oxidation number of Carbon in CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced. The reducing agent is the one which provides electrons by oxidizing itself so in this case; CO is the reducing agent while the C in CO oxidized to produce electrons. </span><span>I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.</span>

6 0

3 years ago

What is the ground state electron configuration for B3+ ?

Arisa [49]

The ground state electron configuration is the arrangement of electrons around the nucleus of an atom with lower energy levels. The electrons occupying the orbitals of varying energy levels naturally falls towards the lowest energy state or ground state.

5 0

2 years ago

A 5 gram round ball has a density of 1.25 grams/milliliter. What is the volume of the round ball?

Arlecino [84]

Explanation:

m=5g

density=1.25g/ml

density=m/v

v=m/density=5/1.25

4 0

3 years ago

If fluorine gas (F₂) occupies a volume of 45.5 L at a pressure of 850

pshichka [43]

Taking into account about the ideal gas law, the mass of fluorine gas is 63.2244 grams.

<h3>What is ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature. The universal constant of ideal gases R has the same value for all gaseous substances.
R is the ideal gas constant.
n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Mass of fluorine gas</h3>

In this case, you know:

P= 850 mmHg= 1.11842 atm (being 1 mmHg= 0.00131579 atm)
V= 45.5 L
T =100 °C= 373 K (being 0°C= 273 K)
R= 0.082 $\frac{atm L}{mol K}$
n= ?

Replacing in the ideal gas law:

1.11842 atm ×45.5 L = n×0.082 $\frac{atm L}{mol K}$ × 373 K

Solving:

n= (1.11842 atm ×45.5 L)÷ (0.082 $\frac{atm L}{mol K}$ × 373 K)

<u><em>n= 1.6638 moles</em></u>

The molar mass of F₂ is 38 g/mole. Then you can apply the folowing rule of three: If by definition of molar mass 1 mole of the compound contains 38 grams, 1.6638 moles of the compound contains how much mass?

$mass= \frac{1.6638 molesx38 grams}{1 mole}$