Ask question

Ask question

All categories

2 years ago

15

A metal rod with a length of 4.66 cm was measured using four different devices. Which of the following measurements is the most

accurate?
4.658 cm
4.635 cm
4.610 cm
4.560 cm

2 answers:

Roman55 [17]2 years ago

5 0

Answer:

4.658 cm

Explanation:

Accuracy is defined as how close the measured value is to a standard value. For example if the given volume of water is 20 ml and the two measured values are 19 ml and 18 ml then the former measured value is more accurate than the later.

Similarly, 4.658 cm is closest to the given length of 4.66 cm.

4.66 - 4.658 = 0.002

4.66 - 4.635 = 0.025

4.66 - 4.610 = 0.05

4.66 - 4.560 = 0.1

iogann1982 [59]2 years ago

3 0

4.658. Accuracy refers to how close the experimental value is to the actual value. Precision is how close a set of data is to one another.

You might be interested in

How many values of ml are allowed for an electron in a 5f subshell??

VladimirAG [237]

There are 7 values of ml are allowed for an electron in a 5f subshell namely: -3 -2, -1, 0, +1, +2, +3

<h3>Further explanation</h3>

In an atom, there are energy levels in the shell and subshell.This energy level is expressed in terms of electron configurations.

Writing the electron configuration starts from the lowest to the highest subshell's energy level. There are 4 sub-shells in an atom's shell, namely s, p, d, and f. The maximum number of electrons for each subshell is

s: 2 electrons
p: 6 electrons
d: 10 electrons and
f: 14 electrons

Electron filling in subshells using the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Each sub-shell also has orbitals drawn in the form of a square box in which there are electrons symbolized by half arrows.

Each orbital in an atom consists of 4 quantum numbers

n is the principal quantum number.
l is the angular momentum / azimuthal quantum number
ml, the magnetic quantum number
ms, the electron-spin quantum number

Value of n: positive integer

value of l: s = 0, p = 1, d = 2, f = 3

ml value: between -l to + l

ms value: +1/2 or -1/2

Determination of electron configurations based on principles:

1. Aufbau: Electrons occupy orbitals of the lowest energy level
2 Hund: electron fills orbitals with the same energy level
3. Pauli: there are no electrons that have 4 equal quantum numbers

So for 5f orbitals, the value of a possible quantum number is

n = 5;

l = 3 (f = 3);

m = -3 -2, -1, 0, +1, +2, +3;

s = + - 1/2

<h3>Learn more</h3>

the locations and properties of two electrons

brainly.com/question/2292596

It's sublevel

brainly.com/question/4520082

a possible full set of quantum numbers

brainly.com/question/5389767

Keywords: orbitals, subshells, quantum numbers

8 0

2 years ago

Read 2 more answers

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0

1 year ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

<u>Step 2:</u> 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 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 years ago

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

5 0

1 year ago

A sodium atom has an electron configuration of 1s22s22p63s1.if the sodium atom becomes ionized, its new electron configuration w

dezoksy [38]

Answer:

Neon

Explanation:

1s² 2s² 2p⁶ 3s¹ or [Ne] 3s¹

The outer most shell is the 3s¹.

For this atom to achieve stability, if it loses the electron in the 3s shell, it would resemble an inert element with a complete octet configuration. Therefore, the atom would be like:

1s² 2s² 2p⁶ which is the configuration of Ne

8 0

2 years ago