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kupik [55]
2 years ago
10

How many rays are needed to be

Chemistry
1 answer:
ollegr [7]2 years ago
8 0

Answer:

two

Explanation:

Every observer would observe the same image location and every light ray would follow the law of reflection. Yet only two of these rays would be needed to determine the image location since it only requires two rays to find the intersection point.

pls name braainly :)

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What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?
Neporo4naja [7]

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

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3 years ago
One difference between d-glucose and l-glucose is
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<span>only D-glucose is found in disaccharides and polysaccharides.</span>
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Which statement is completely accurate?
AURORKA [14]

Answer:The same chemical element can have a different number of neutrons and still be the same element. We refer to the atoms of the same element with different numbers of neutrons as "isotopes"

Explanation:

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2 years ago
What is it called when copper can be stretched into s thin wire?
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Its called ductility
6 0
3 years ago
2) 2KClO3 --&gt; 2KCl + 3O2
aleksandrvk [35]

2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2

a)

2 \text{ mols of KClO}_3 \equiv 3  \text{ mols of O}_2

19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5  \text{ mols of O}_2

\boxed{19 \text{ mols of KClO}_3 \equiv 28,5  \text{ mols of O}_2}

b)

2 \text{ mols of KClO}_3 \equiv 2  \text{ mols of KCl}

62 \text{ mol of KClO}_3 \equiv 62  \text{ mol of KCl}

Using the atomic mass given in the periodic table:

62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

62\cdot122,5 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

7595 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

\boxed{7,595 \text{ kg of KClO}_3 \equiv 62  \text{ mol of KCl}}

c)

2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3

3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}

Using the atomic mass given in the periodic table:

3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5)  \text{ g of KCl}

96\text{ g of O}_2 \equiv 149\text{ g of KCl}

\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}

\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}

This result is an aproximation.

8 0
3 years ago
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