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aniked [119]
4 years ago
14

Determine the volume of a piece of gold with a mass of 318.97g

Chemistry
1 answer:
djyliett [7]4 years ago
6 0

Answer:

318 g / 19.32 g     v = 16. your volume is 16 hope this helps

Explanation:

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Why are the concentrations of [H3O+] and [OH−] equal in pure water?
kotegsom [21]

Answer : The correct option is, (A) [H_3O^+]=[OH^-] because one of each is produced every time an [H^+] transfers from one water molecule to another.

Explanation :

As we know that, when the two water molecule combine to produced hydronium ion and hydroxide ion.

The balanced reaction will be:

HOH+HOH\rightarrow H_3O^++OH^-

Acid : It is a substance that donates hydrogen ion when dissolved in water.

Base : It is a substance that accepts hydrogen ion when dissolved in water.

From this we conclude that, the hydrogen ion are transferred from one water molecule to the another water molecule to form hydronium ion and hydroxide ion. In this reaction, one water molecule will act as a base and another water molecule will act as an acid.

Hence, the correct option is, (A)

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4 years ago
What is the name of the structure formed when fertilization occurs?
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Answer:

zygote

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3 0
3 years ago
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The radius of an atom is closest in size to a
Andreas93 [3]

Answer:

Explanation:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons. Since the boundary is not a well-defined physical entity, there are various non-equivalent definitions of atomic radius. Three widely used definitions of atomic radius are: Van der Waals radius, ionic radius, and covalent radius.

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3 years ago
A variable parameter
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Answer

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4 0
3 years ago
Given the following reaction: H2(g)+I2(s) → 2HI(g) with a ∆H of 52.9 kJ. What is the change in enthalpy for the following reacti
NeTakaya

<u>Answer:</u> The change in enthalpy of the reaction is -26.45 kJ

<u>Explanation:</u>

The given chemical reaction follows:

H_2(g)+I_2(s)\rightarrow 2HI(g);\Delta H=52.9kJ

The enthalpy change of the reaction when 2 moles of HI is formed is 52.9 kJ

Now, the chemical reaction for which enthalpy change is to be calculated is:

HI(g)\rightarrow \frac{1}{2}H_2(g)+\frac{1}{2}I_2(s)

Sign for enthalpy change will be reversed, because the reaction is reverse.

By Stoichiometry of the reaction:

When 2 moles of Hi is reacted, the enthalpy change of the reaction is -52.9 kJ

So, when 1 mole of HI will be reacted, the enthalpy change of the reaction will be = \frac{-52.9}{2}\times 1=-26.45kJ

Hence, the change in enthalpy of the reaction is -26.45 kJ

5 0
3 years ago
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