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lubasha [3.4K]
3 years ago
6

How To Divide In Math Easily?Please Help My 6 Yrs OldQuestion ​

Chemistry
2 answers:
mr_godi [17]3 years ago
7 0

Answer:

see down

Hope this help

<3

<h2><u><em>Red</em></u></h2>

viktelen [127]3 years ago
6 0

Answer:

see the example that you can do

ask me anything

Hope this helps

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Can somebody help me, please!
zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 72/100= 0.72      or        72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

  • mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 28/100= 0.28       or        28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

  • mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

<u>Atomic Mass of Rubidium:</u>

Add the two numbers together.

  • Rb-85 (61.2) and Rb-87 (24.36)
  • 61.2+24.36=85.56 amu
4 0
3 years ago
Determine the empirical formula of a
k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold  = 2.6444g

Mass of Chlorine  = 0.476g

Unknown:

Empirical formula  = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements                                     Au                                             Cl

Mass                                         2.6444                                     0.476

Molar mass                                 197                                          35.5

Number of moles                  2.6444/197                                 0.476/35.5

                                                 0.013                                           0.013

Divide by the

smallest                                 0.013/0.013                                 0.013/0.013

                                                       1                                                   1

The empirical formula of the compound is AuCl

3 0
2 years ago
How can we prevent rusting of iron??
dimaraw [331]

water plus oxygen equals rust so keep water away from the iron to prevent rusting or dry the iron off then apply alchohol to cleanse it

5 0
3 years ago
How many moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl? ​
Strike441 [17]

Answer:

0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl

Explanation:

This is the reaction:

2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)

To make 3 moles of H₂, we need 2 moles of Al.

By conditions given, we will find out how many moles of H₂ do we have.

Let's use the Ideal Gas Law

P. V = n . R . T

1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K

(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n

0.182 mol = n

So the rule of three will be:

If 3 moles of H₂ came from 2 moles of Al

0.182 moles of H₂ will come from x

(0.182 .2) / 3 = 0.121 moles

4 0
3 years ago
What mass of cu(s) is electroplated by running 26.5 a of current through a cu2+(aq) solution for 4.00 h?
kramer
T = 14400 s 
26.5 x 14400=381600 C 
381600/96500=3.95 Faradays 
Cu2+ + 2e- = Cu 
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g 

moles Au = 33.1 / 196.967 g/mol=0.168 
Au+ + 1e- = Au 
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday 
0.168 x 96500=16217 Coulombs 
16217 / 5.00=3243 s => 54 min
7 0
3 years ago
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