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2 years ago

What is the volume of an object with the mass of 7.9 grams in the density of 2.28g/ml.

Chemistry

1 answer:

Schach [20]2 years ago

7 0

Answer:

3.7mL is the volume of the object

Explanation:

To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:

7.9g * (1mL / 2.28mL) =

<h3>3.7mL is the volume of the object</h3>

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A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick u

Butoxors [25]

Answer:

One of the main uses of the cathode ray tube is in the Cathode ray oscilloscope

Explanation:

Cathode rays are produced when a gas in an evacuated glass at very low pressure and high pressure breaks up into positive and negative ions. the negative ions move towards the anode(positive electrode) while the positive ions move towards the cathode(negative electrode), and there they knock off electrons (which are known as cathode rays) from the metal plate of the cathode.

Cathode ray tubes are mainly used in oscilloscopes, television picture tubes and in computer screens.

The cathode ray oscilloscope is used in a.c. and d.c. voltage measurements, observation of waveforms, time measurements, etc.

5 0

3 years ago

A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate

inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

water content of sample is taken as;

w = (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 / 1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0

3 years ago

A gold coin contains 3.47 × 10^23<br> gold atoms.<br> What is the mass of the coin in grams?

VashaNatasha [74]

Answer:

Mass = 114.26 g

Explanation:

Given data:

Number of gold atoms = 3.47×10²³ atoms

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms

0.58 mol

Mass of gold:

Mass = number of moles × molar mass

Mass = 0.58 mol × 197 g/mol

Mass = 114.26 g

4 0

3 years ago

Please help me and I’ll give you brainiest!!!!!!!!! <br><br><br><br> Have a wonderful day!!:)))

Step2247 [10]

Answer:

up down up down thats the pattern

Explanation:

5 0

2 years ago

Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown

defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

Divide by the smallest mol(Mn=0.0115)

Mn : O =

$\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2$

The empirical formula : MnO₂

8 0

2 years ago