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1 year ago

A length of glass tubing is 0.525 m. How many inches long is the tubing? (2.54 cm = 1 inch)

Chemistry

2 answers:

lara31 [8.8K]1 year ago

6 0

If, (.525m)(100 cm/m)(1/2.54 in/cm) = 20.7 in or C

valkas [14]1 year ago

5 0

Answer : The correct option is, (C) 20.7 inches

Explanation :

The conversion used from meter to centimeter is:

1 m = 100 cm

The conversion used from centimeter to inches is:

2.54 cm = 1 inch

First we have to convert the length of glass tubing from meter to centimeter.

As, $1m=100cm$

So, $0.525m=\frac{0.525m}{1m}\times 100cm=52.5cm$

Now we have to convert the length of glass tubing from centimeter to inches.

As, $2.54cm=1\text{ inch}$

So, $52.5cm=\frac{52.5cm}{2.54cm}\times 1\text{ inch}=20.7\text{ inches}$

Therefore, the length of glass tubing in inches are 20.7 inches.

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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

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1 year ago

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A sample of neon effuses from a container in 72 seconds. the same amount of an unknown noble gas requires 147 seconds. you may w

Alexxx [7]

The second gas is identified as follows

by Graham law formula
let the unknown gas be represented by letter y

=time of effusion of Neon/ time of effusion of y = sqrt (molar mass of neon/molar mass of y)

= 72 sec/ 147 sec = sqrt( 20.18 g/mol/ y g/mol)

square the both side to remove the square root sign

72^2/147^2 = 20.18 g/mol/y g/mol

=0.24 = 20.18g/mol/y g/mol

multiply both side by y g/mol
= 0.24 y g/mol = 20.18g/mol
divide both side by 0.24
y = 84 g/mol

y is therefore Krypton since it is the one with a molar mass of 84 g/mol

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1 year ago

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Which element has 32 protons in its nucleus? phosphorus cobalt germanium sulfur

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Germanium is the element that has 32 protons in its nucleus.

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2 years ago

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How many grams of C5H12 must be burned to heat 1.39 kg of water from 21.2 °C to 97.0 °C? Assume that all the heat released durin

faust18 [17]

Answer:

m = 8.9856 g

Explanation:

In order to do this, we need to write the expressions that are to be used. First, to calculate heat:

Q = m*C*ΔT (1)

Where C would be heat capacity of the substance.

The heat can also be relationed with the moles and enthalpy of a compound using the following expression:

Q = n*ΔH (2)

Finally for the mass of any compound, we use the following expression:

m = n*MM (3)

So, in order to calculate the grams of pentane (C5H12), we need to calculate the moles of the compound, and to do that, we need the heat exerted.

So, as we are using water, let's calculate the heat that is been exerted with the water. The C of the water is 4.186 J/g °C so:

Q = (1.39 * 1000) * 4.186 * (21.2 - 97)

Q = -441,045.33 J

This is the heat neccesary to burn pentane and heat water. Now, with this value, let's calculate the moles used of pentane with expression (2). The ΔH of the pentane is -3,535 045.kJ/mol or -3.535x10⁶ J/mol. Solving for n we have:

n = -441,045.3 / -3.535x10⁶

n = 0.1248 moles

Finally, we can calculate the grams needed with expression (3). The molar mass of pentane is 72 g/mol

m = 0.1248 * 72

m = 8.9856 g

This is the mass needed to heat 1.39 kg of water

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2 years ago

Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.

Lelu [443]

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.

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1 year ago