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GarryVolchara [31]
3 years ago
9

Can anyone please help??

Chemistry
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

Don't know that one

Explanation:

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How many moles are in 1.806 x 1024 molecules of bromine?
Pepsi [2]
<h3>Answer:</h3>

2.999 mol Br

<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>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.806 × 10²⁴ molecules Br

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

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 1.806 \cdot 10^{24} \ molecules \ Br(\frac{1 \ mol \ Br}{6.022 \cdot 10^{23} \ molecules \ Br} ) = 2.999 mol Br

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

Our final answer is already in 4 sig figs, so there is no need to round.

7 0
3 years ago
Read 2 more answers
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

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3% icebergs, lakes, rivers, and ground water
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BaS + PtF2 = BaF2 + PtS
dedylja [7]
I think it is 
Barium Sulfide + Platinum Difluoride = Barium Fluoride + Platinum Sulfide<span>
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Sue dissolved a certain amount of salt in 400 grams of water to obtain 405 grams of salt solution. What was the mass of the salt
svp [43]
The answer would be 5 grams of salt
6 0
3 years ago
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