How many atoms of oxygen are in 75.6g of silver nitrate (AgNO3)?​

The chemical equation can be expressed as:

2H2 + O2 = 2H2O

Given the amounts of the reactants, we need to identify the limiting reactant before calculating the amount of product generated.

4.0 g H2 ( 1 mol / 2.02 g ) = 1.98 mol H2
5.0 g O2 ( 1 mol / 32 g ) = 0.1563 mol O2

The limiting reactant is O2, as it will be fully consumed in the reaction.

0.1563 mol O2 ( 2 mol H2O / 1 mol O2 ) ( 18.02 g / mol ) = 5.6 g H2O will be produced

Response:

The specific heat of the alloy

Clarification:

Weight of the alloy = 25 gm

Initial temperature = 100°c = 373 K

Weight of the water = 90 gm

Initial temperature of water = 25.32 °c = 298.32 K

Final temperature = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

[[ - ] = ( - )

25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This gives us the specific heat of the alloy.

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Response: sodium + water → sodium hydroxide + hydrogen

Clarification:

1) The initial part of the statement "sodium combines with water" indicates that these pure substances serve as the reactants.

Sodium belongs to the alkali metals (group 1: Li, Na, K, Rb, Cs, Fr).

A common characteristic of alkali metals is their reactive behavior with water, resulting in the formation of the respective hydroxide and hydrogen gas.

2) The following words, "to produce" suggest that what comes next represents the products of the reaction. In chemical equations, this is indicated by the arrow →

3) Finally, it specifies the products of the reaction: sodium hydroxide and hydrogen gas.

4) The word equation observes the sequence reactants → products, translates here as sodium + water → sodium hydroxide + hydrogen

5) Using chemical symbols and formats, this can be represented as:

Na(s) + H₂O (l) → NaOH(aq) + H₂(g).