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2 years ago

How many significant figures in 8400

Chemistry

2 answers:

IrinaK [193]2 years ago

5 0

Answer:

there are two significant figures is the number 8400

Explanation:

Varvara68 [4.7K]2 years ago

4 0

There are two significant figures in the number 8400

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1. How many grams would 8.1 x 1021 molecules of sucrose (C12H22011) weigh?

Nimfa-mama [501]

Answer:

Mass = 4.6 g

Explanation:

Given data:

Number of molecules of sucrose = 8.1 ×10²¹ molecules

Mass of sucrose = ?

Solution:

First of all we will calculate the number of moles by using Avogadro number.

1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules

1.35 × 10⁻² mol

Mass of sucrose:

Mass = number of moles × molar mass

Molar mass = 342.3 g/mol

Mass = 1.35 × 10⁻² mol ×342.3 g/mol

Mass = 462.1 × 10⁻² g

Mass = 4.6 g

7 0

3 years ago

Which of the following best describes a mole as used in chemistry?

bazaltina [42]

Answer: a very large number of objects

Explanation: Mole is the amount of objects and it has the usually Avogadro number of atoms, molecules, ions, etc in chemistry

8 0

2 years ago

What does composing mean?

levacccp [35]

Answer:

$Composing \: means \: to \: come \: together \: to \: \\ form \: or \: make \: something.$

Explanation:

Thank you!!!!

4 0

2 years ago

Read 2 more answers

I need help please!!! Name the parts making up the female part of the flower and state what each part does. Please use your own

Greeley [361]

<h2>Answer:-</h2>

The female part is the pistil. The pistil usually is located in the center of the flower and is made up of three parts: the stigma, style, and ovary. The stigma is the sticky knob at the top of the pistil. It is attached to the long, tubelike structure called the style.

6 0

2 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$