If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)

1) Chromium(III) nitrate is classified as acidic because it is derived from a weak base (chromium(III) hydroxide Cr(OH)₃) and a strong acid (nitric acid HNO₃). 2) Sodium hydrosulfide is considered basic because it results from a strong base (sodium hydroxide NaOH) and a weak acid (hydrogen sulfide H₂S). 3) Zinc acetate is slightly basic since zinc hydroxide (Zn(OH)₂) is a stronger base than acetic acid (CH₃COOH).

The balanced chemical equation for the neutralization of HCl with is:

Given weight of = 5g

Moles of =

Volume of HCl solution =

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2

The enthalpy of neutralization per mole of

=

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).

Approximately 400 mL of a 5.0% glucose solution.

Response:

1.98 M

Clarification:

Provided data

• Starting volume (V₁): 93.2 mL

• Starting concentration (C₁): 2.03 M

• Water volume added: 3.92 L

Step 1: Convert V₁ to liters

Using the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)

The final volume is the total of the initial volume and the added water volume.

Step 3: Calculate the final concentration (C₂)

Utilizing the dilution rule.