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3 years ago

What is the correct formula for the compound formed between sodium and iodine based on their positions in the periodic table?

1 answer:

6 0

Sodium (Na) has a +1 charge and Iodine ( I ) has a -1 charge. To create a molecule of sodium iodide the charges will need to balance.

Because the charges on anion and cation are the same; the molecular formula will be NaI

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jok3333 [9.3K]

Answer:

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Explanation:

im top global chou in ml

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3 years ago

The electron configurations of two unknown elements X and Y are shown . X: 1s^ 2 2s^ 2 2p^ 6 3s^ 1 Y: 1s^ 2 2s^ 1 Which statemen

PIT_PIT [208]

Answer:

See below.

Explanation:

They both have 1 valency electron so will be metallic and in the same Group (Group 1) of the Periodic Table, so will have similar properties.

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3 years ago

Read 2 more answers

The percent yield of this reaction is consistently 92.0%. CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g) How many grams of sulfur would be

dybincka [34]

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .

since ,

Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂

Since ,

the Molecular weight of CS₂ = 76

Given , mass of CS₂ = 72.57 g

Moles of CS₂ = 72.57 / 76 = 0.95 mol

Since ,

The yield is 92.0 % .

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g .

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3 years ago

Hello! I have a green pigment called the chlorophyll. I also absorb solar

vodka [1.7K]

Answer:

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Explanation:

4 0

2 years ago

A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4

natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

$M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M$

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3 years ago