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3 years ago

13

The density of gold is 19.3 grams per cubic centimeter. What is the mass of a bar of gold that measures 15.24 cm in length, 10.1

6 cm in width, and 5.08 cm in height?

1 answer:

kumpel [21]3 years ago

4 0

Answer:

40.75

Explanation:

15.24 × 5.08 × 10.16 =786.579072

786.579072/19.3=40.75

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based on the techniqes yo have learned in the organic chemistry lab how would you seperate any unreated alcohol from the ester

mihalych1998 [28]

Answer:

Fractional distillation and HP-LC

Explanation:

This is a technique useful for analytes with close boiling points. Any alcohol-ester azotopes can be further refined using high-performance liquid chromatography (HP-LC) column.

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3 years ago

Describe the differences among primitive, igneous, sedimentary, and metamorphic rock, and relate these differences to their orig

Akimi4 [234]

Answer:

Rocks are the aggregate of minerals. There are three distinct categories of rocks, namely the sedimentary, metamorphic and the igneous rocks.

The sedimentary rocks are formed from the deposition, compaction, and lithification of soft sediments that are transported from one place to another by the agents such as wind, water, and ice. For example, Sandstone and Shale.

The metamorphic rocks are derived from the previously existing sedimentary, igneous or other metamorphic rocks, due to the influence of extremely high pressure as well as temperature conditions. For example, Quartzite and Marble.

The igneous rocks are those rocks that are formed from the cooling and crystallization of magma. For example, Granite and Diorite.

All these three types of rocks are formed by different processes and their mode of origins are also different.

5 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago

The mass number of an isotope of uranium is 238, and its atomic number is 92. Which symbol best represents this isotope of urani

laila [671]

$238_{U}$

That is a capital U with the number 238 at upper left of the letter

5 0

3 years ago

Read 2 more answers

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spheric

Luba_88 [7]

Complete Question

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

Answer:

The pressure required is $P_2= 12.2 \ atm$

Explanation:

Generally the volume of a sphere is mathematically denoted as

$V_s = \frac{4}{3} * \pi r^3$

Substituting $r = \frac{d}{2} = \frac{74}{2} = 37cm$

$V_s = \frac{4}{3} * 3.42 * (37)^2$

$V_s = 2.121746 *10^5 cm^3$

Converting to Liters

$V_s = \frac{2.121746 *10^5}{1000}$

$V_s= 212.1746L$

Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L

So

From Charles law

$P_1V_1 = P_2 V_s$

Substituting $V_1 =2600 L$ , $P_1 = 1 atm$ , $V_s =212.1746L$ , and making $P_2$ the subject we have

$P_2 = \frac{P_1 * V_1}{V_s}$

$= \frac{1 * 2600}{212.1746}$

$P_2= 12.2 atm$

6 0

3 years ago