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3 years ago

What is the chemical name for the compound with the formula Na2S?

Chemistry

2 answers:

Svet_ta [14]3 years ago

7 0

Answer : The correct option is, sodium sulfide

Explanation :

The given formula is, $Na_2S$

The rules for the naming of the binary molecular compound are :

First element in the formula is named first and keep its element name.
Gets a prefix if there is a subscript on it.
Second element is named second.
Use the root of the element name plus suffix (-ide).
Always use a prefix on the second element.

In the given formula, the first element is (Na) sodium and the second element is sulfur. Thus, the chemical name of the compound is, sodium sulfide.

Therefore, the chemical name for the compound with the formula $Na_2S$ is, sodium sulfide.

Marina86 [1]3 years ago

3 0

I think it's Sodium Sulfide, because Na2S contains one Sodium and two Sulfur. Plus Sulfide<span> is used to describe any of three types of chemical compounds that contain sulfur.</span>

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2 years ago

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3 years ago

If a force of 1.4N is applied to a block with a mass of 7kg, what is the acceleration of the block in meters per second squared?

Anton [14]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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We have the final answer as

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2 years ago

Read 2 more answers

What is the concentration of OH − and pOH in a 0.00066 M solution of Ba ( OH ) 2 at 25 ∘ C? Assume complete dissociation.

Allushta [10]

<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

<u>Explanation:</u>

We are given:

Concentration of barium hydroxide = 0.00066 M

The chemical equation for the dissociation of barium hydroxide follows:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M$

Putting values in above equation, we get:

$pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88$

Hence, the hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

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3 years ago

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Trava [24]

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3 years ago