All categories

3 years ago

How many ne atoms are contained in 32.0g of the element?

Chemistry

1 answer:

soldi70 [24.7K]3 years ago

7 0

(32.0 g Ne) / (20.1797 g Ne/mol) × (6.022 × 10^23 atoms/mol) = 9.55 × 10^23 atoms N3

You might be interested in

Suppose you wanted to monitor a pH change between 5.3 and 6.3. Which indicator would be most appropriate?

umka2103 [35]

It would be acidic based indicator.
0-6 is acidic
7 is neutral
8-14 is alkaline

7 0

3 years ago

Read 2 more answers

What is the formula (with cation and anion charges) for copper (II) hydroxide?

Thepotemich [5.8K]

Answer:

Cu²⁺(OH)⁻₂

Explanation:

Step 1: Find the symbols for the substances

For copper, see your periodic table. Copper is "Cu".

For hydroxide, see your polyatomic ions chart. Hydroxide is "OH".

Step 2: Find the charges for each element.

Copper (II)'s charge is 2, which is the roman numerals in the brackets.

Hydroxide's charge is -1.

Charges are written as superscripts. You do not need to write "1".

In formulas, write the positively charged atom first.

Cu²⁺(OH)⁻ Put brackets around hydroxide because it has 2 elements.

Step 3: Write the number of atoms.

The number of atoms that an element has is the same as its partner's charge.

The charge of copper is 2, so hydroxide has 2 atoms.

The charge of hydroxide is 1, so copper has 1 atom.

Cu²⁺₁(OH)⁻₂

You do not need to write the "1" for atoms.

Cu²⁺(OH)⁻₂

8 0

3 years ago

What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?

Vladimir79 [104]

Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2 -------> 32 g O2 -----------> 36 g H2O
↓ ↓ ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.

4 0

3 years ago

How do I do question number 2?

ale4655 [162]

Answer:

I cant help

Explanation:

4 0

3 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

3 years ago