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2 years ago

A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3 . calculate the density of gold.

Chemistry

1 answer:

Fantom [35]2 years ago

6 0

Answer:

19.29 g.cm⁻³

Solution:

Data Given:

Mass = 301 g

Volume = 15.6 cm³

Formula Used:

Density = Mass ÷ Volume

Putting values,

Density = 301 g ÷ 15.6 cm³

Density = 19.29 g.cm⁻³

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Convert 6 grams of carbon atoms into miles

anyanavicka [17]

How many grams Carbon in 1 moles? The answer is 12.0107.
We assume you are converting between grams Carbon and mole.
You can view more details on each measurement unit:
molecular weight of Carbon or moles
The molecular formula for Carbon is C.
The SI base unit for amount of substance is the mole.
1 grams Carbon is equal to 0.0832590939745 mole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams Carbon and mole.
Type in your own numbers in the form to convert the units!

4 0

2 years ago

If the experiment were repeated using a cuvettes with a path length of 2.0 cm, how would that affect the absorbance values for t

inessss [21]

Answer:

..............................

5 0

2 years ago

For what reasons would modern scientists accept dalton's atomic theory but not the idea of an atom proposed by the greek philoso

soldi70 [24.7K]

Solution:

The idea of an atom proposed by the Greek philosophers because:

According to Greek Philosopher, matter is composed of small and indivisible particles called atoms. He introduced atoms as too small to be seen, unchangeable, completely solid without internal structure. He proposed that atoms are of variety of shapes and sizes which is responsible for different types of matter.

But according to Dalton’s atomic theory, chemical elements have atoms, which are identical in weight. The different elements have different atoms of different weight. Atoms can combine in whole-number ratios to form compounds. These observations are already introduced by Greek philosopher, but the idea of atomic weight is introduced by Dalton. He introduced the list of 21 elements with their atomic weights and, he was the first to propose the element’s symbol.

4 0

2 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!

6 0

2 years ago

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

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

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

The volume of the gas is 2.80 L.

7 0

2 years ago