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Oxana [17]
2 years ago
9

2. Calculate the density of a metal that occupies 17.75 cm and has a mass of 342.93 g. [D = m/V]

Chemistry
1 answer:
satela [25.4K]2 years ago
3 0

Answer:

19.32

Explanation:

the density is given by the mass over volume

the mass In this case is 342.93 and the volume is 17.75

d=342.93g/17.75cm

=19.32g/cm

i hope this helps

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Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an
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Answer: the correct option is 2, 3, 4, 1.

Explanation:alphabetic indexing order is the order in which files or names are being arranged according to the alphabet. In the following names:

2) KST Enterprises

3)Leland and Son Graphics

4)Lucinda Topper

1) Topper & Casey Plumbing.

While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

5 0
2 years ago
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NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a
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Answer:

λ = 0.0167 m = 16.7 mm

Explanation:

The wavelength of these radio waves can be found out by using the formula for the speed of radio waves:

v = fλ

where,

v = speed of radio waves = speed of light = 3 x 10⁸ m/s

f = frequency of radio waves = 18 GHz = 18 x 10⁹ Hz

λ = Wavelength = ?

Therefore,

3 x 10⁸ m/s = (18 x 10⁹ Hz)λ

λ = (3 x 10⁸ m/s)/(18 x 10⁹ Hz)

<u>λ = 0.0167 m = 16.7 mm</u>

4 0
2 years ago
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
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Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL

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3 years ago
Using the periodic table, determine the ion charges of the following families of elements if valence electrons were removed or a
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Group I=1+

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6 0
2 years ago
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8.0 mol AgNO3 reacts with 5.0 mol Zn in
Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • AgNO₃: 2 moles
  • Zn: 1 mole
  • Ag: 2 moles
  • Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

#SPJ1

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