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Sholpan [36]
3 years ago
6

What volume of 0.160 m li2s solution is required to completely react with 130 ml of 0.160 m co no3 2?

Chemistry
1 answer:
Over [174]3 years ago
3 0
The  volume  of  0.160    m   Li2S  solution  required  to  completely  react  with  130 ml  of 0.160  CO(NO3)2  is calculated   as  below

write the  reacting  equation

Co(NO3)2 +  Li2S = 2LiNO3  +  COS

find the    moles  of CO(NO3)2  = molarity  x  volume

=  130 ml  x  0.160=20.8  moles

since the reacting moles between CO(NO3)2  to LiS  is   1:1  the  moles of LiS  is  also  20.8  moles

volume  of Lis  is  therefore =  moles of Lis/ molarity  of LiS

=  20.8/0.160 =  130 Ml
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Diano4ka-milaya [45]

Answer:

your answer is b and hope ur help and mark me brainlist

6 0
3 years ago
How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *
yuradex [85]
<h3>Answer:</h3>

\displaystyle 0.1 \ mol \ Fe

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.022 × 10²² atoms Fe (iron)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})
  2. Divide:                    \displaystyle 0.1 \ mol \ Fe
7 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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5 0
2 years ago
I have no idea how to do this help plz!
Wittaler [7]
The answes are
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8 0
3 years ago
Read 2 more answers
6 How many moles are in a 321 g sample of magnesium?
Tatiana [17]

Answer:

13.4mol of Mg

Explanation:

Given parameters:

Mass of magnesium = 321g

Unknown:

Number of moles  = ?

Solution:

The number of moles of a substance is given as;

  Number of moles  = \frac{mass}{molar mass}  

 Molar mass of Mg = 24g/mol

 Insert the parameters and solve;

        Number of moles  = \frac{321}{24}   = 13.4mol of Mg

3 0
3 years ago
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