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spin [16.1K]
3 years ago
6

2H20 → 2H2 + O2 Type of reaction?

Chemistry
1 answer:
snow_tiger [21]3 years ago
3 0

Answer:

  • Notice how the elements split
  • It is a <em>decomposition</em><em> </em><em>reaction</em><em> </em>
  • I do exams and quizzes if ur interested
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What scientists do that is the basis for their investigations
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<span>The basis of any scientific investigation is a hypothesis and a research question. These guide your research and help you focus on an area that is narrow enough to study for meaningful results. The research question helps quantify how you are going to study the hypothesis.</span>
7 0
3 years ago
What is the total number of moles of solute contained in 0.5 liter of 3.0 m
myrzilka [38]

Based on science 4) 1) 0.25M  

2) 0.66M  

3) 1.5M  

4) 4.0M ––> Therefore, this is the most concentrated  

5) 30g NaOH(1 mol NaOH / 40gNaOH) = 0.75 mol NaOH / 0.500L = 1.5M NaOH (3)  

6) 1) 1.0 mol / L(1 L) = 1 mol H2SO4  

2) 1 mol / L (2 L) = 2 mol H2SO4  

3) 0.50 mol / L ( 1.0L) = 0.50 mol H2SO4 ––>

7) 0.200 mol / L (1 L)(74.6g / mol) = 14.92g (2)  

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This is the answer  : 1m

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3 years ago
Why series fatty acid carbons seen in double from?
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8 0
2 years ago
What does fire mean​
Bond [772]

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3 0
3 years ago
A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man
kap26 [50]
<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

  • Mass of NaBr sample is 11.97 g
  • % composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

  • Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

                         = 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0
3 years ago
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