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BabaBlast [244]

3 years ago

11

For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?

*
4 points
magnesium
oxygen
magnesium oxide
dimagnesium dioxide

1 answer:

Anit [1.1K]3 years ago

3 0

Explanation:

hope this can help youuu

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The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.9% H, and 31.4% O. What’s the empirical

blsea [12.9K]

To find the empirical formula you would first need to find the moles of each element:

58.8g/ 12.0g = 4.9 mol C

9.9g/ 1.0g = 9.9 mol H

31.4g/ 16.0g = 1.96 O

Then you divide by the smallest number of moles of each:

4.9/1.96 = 2.5

9.9/1.96 = 6

1.96/1.96 = 1

Since there is 2.5, you find the least number that makes each moles a whole number which is 2.

So the empirical formula is C5H12O2.

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3 years ago

Find the number of Grams

zaharov [31]

Answer: 462 g

Explanation:molar mass is M= 63.55 +2·(12.01+14.01)= 115.59 g/mol.

Mass m= n·M = 4.0 mol·115.59 g/mol= 462.36 g

8 0

2 years ago

Which substances will make a salt when combined?

babymother [125]

The answer is D. Fertilizer and vinegar

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3 years ago

Read 2 more answers

Which element, represented by X, reacts with oxygen to produce the compound X2O?

natali 33 [55]

Answer:

lithium

Explanation:

this is because lithium has a valency of 1 and oxygen has a valency of 2 thereby exchanging valency to create Li²0

7 0

3 years ago

The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad

Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = $x\times \frac{56.4}{100}=(0.564x)pm$

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

$2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}$

$2\times x+2\times 0.564x=566$

$2x+1.128x=566$

$3.128x=566$

$x=180.9\approx 181pm$

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

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3 years ago