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FrozenT [24]
3 years ago
6

If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r

esults? how?
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
6 0
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with 
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
OlgaM077 [116]3 years ago
6 0

If the standard solutions had unknowingly been made up to be 0.0024 M AgNO₃ and 0.0040 M K₂CrO₄ the chemical equation would be:  

2AgNO₃ + K₂CrO₄ → Ag₂CrO₄ (s) + 2KNO₃

By mixing 0.0024 M AgNO₃ and 0.004 M K₂CrO₄ one would have Ag₂CrO₄ precipitated out.  

Thus, one is left with 0.0024 M KNO₃ mixed with (0.004 -0.0024 / 2) M = 0.0028 M of K₂CrO₄.  

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B. Developing theories using many lines of evidence

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Why do you think it is important to use the same blanace throughout the entire experiment?
Zepler [3.9K]

It is important to use the same balance throughout the entire experiment since the calibration of each balance is not the same and changing balances could result in a systematic error.

There are three types of errors that could affect the results of the experiment. The effect of random or indeterminate errors is hard to predict, its effect on the results of the experiment could be different every time. The second type of error is the systematic or determinate error, which causes a shift in results in a specific direction. The last type of error in an experiment is human error.

The type of error that could be related to the use of different balances throughout the experiment is the systematic error. Instruments could be a source of error especially if they are poorly calibrated. Also, analytical balances are calibrated differently which may result in inaccuracy in the weighing of chemicals.

To learn more, please refer to brainly.com/question/11541675.

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2 years ago
A company sells a natural medicine with plant ingredients to help treat dementia, a disease in which people have a hard time rem
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4 years ago
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In a 3.7- gram sample, there are______ atoms of gold
beks73 [17]
<h3>Answer:</h3>

1.1 × 10²² atoms Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<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
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.7 g Au

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

Avogadro's Number

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

<u />3.7 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au} )(\frac{6.022 \cdot 10^{23} \ atoms \ Au}{1 \ mol \ Au} ) = 1.13121 × 10²² atoms Au

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au

7 0
3 years ago
You burn 270 grams of glucose during respiration. How
BlackZzzverrR [31]

Answer:

396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.

Explanation:

To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;

C6H12O6 + 6O2 -------> 6CO2 + 6H2O

1 mole of C6H12O6 reacts to form 6 moles of CO2

Then, calculate the molar mass of the two variables;

Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol

Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol

Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles

1* 180 g of glucose yields 6 * 44 g of CO2

180 g of glucose = 264 g of CO2

If 270 g of glucose were to be used, how many grams of CO2 will be produced;

so therefore,

180 g of glucose = 264 g of CO2

270 g of glucose = x grams of CO2

x = 264 * 270 / 180

x = 71 280 / 180

x = 396 g of CO2.

In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.

8 0
3 years ago
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