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4 years ago

What would be an example of a chemical change that took a long time to occur?

Chemistry

1 answer:

nydimaria [60]4 years ago

5 0

<span>rusting takes a long time, and rusting is a chemical change.</span>

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1. Find the molarity of 25g of sodium hydroxide dissolved in 1L of water. Identify the solute and solvent in this solution.

wolverine [178]

Answer:

.6250 M

The Sodium Hydroxide is the solute, and the water is the solvent.

Explanation:

mol/volume= Molarity

(25/22.99+16+1.008)/1=.6250 M.

6 0

3 years ago

A runner wants to run 12.8 km . She knows that her running pace is 7.1 mi/h .

Alexus [3.1K]

<span>67.2 minutes

First convert the runner's speed from mi/h to km/h by multiplying by 1.60934 km/mi, giving 7.1 mi/h * 1.60934 km/mi = 11.43 km/h Divide the distance by her speed. 12.8 km / (11.43 km/h) = 1.12 h Convert hours to minutes by multiplying by 60 1.12 h * 60 min/h = 67.2 min</span>

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4 years ago

1) Which New York State landscape region is located at 42 N, 75 W?

denis23 [38]

Answer:C) the Catskills

Explanation:

8 0

3 years ago

What does a high melting point temperature indicate about a substance?

Lilit [14]

They have strong intermolecular forces between their atoms, hence why more heat is necessary.

3 0

3 years ago

Read 2 more answers

Hydrogen sulfate reacts with water to produce sulfate ions and hydronium ions: hso4−+h2o⇌so42−+h3o+ identify the conjugate acid-

Angelina_Jolie [31]

The generic equation for a reaction between an acid and water is

$HX_{(aq)} + H_{2}O_{(aq)} --> X_{(aq)}^{-} + H_{3}O_{(aq)}^+$

When an acid "reacts" with water, water acts as the base that accepts the proton (H+) from the acid. The remaining ion that is formed after the acid has donated its proton is called the conjugate base ( $X^-$ ), and the conjugate acid-base pair is $HX$ - $X^-$ .

Hydrogen sulfate ( $HSO_{4}^-$ ) is an ion from sulfuric acid. It is still an acid in itself and can "react" with water (( $H_{2}O$ ) to form the sulfate ( $SO_{4}^2-$ ) and hydronium ( $H_{3}O^+$ )ions.

$HSO_{4(aq)}^{-} + H_{2}O_{(aq)} --> SO_{4(aq)}^{2-} + H_{3}O_{(aq)}^+$

Based on the previous discussion, $SO_{4(aq)}^2-$ is identified to be the conjugate of the acid $HSO_{4(aq)}^-$ . Thus, the conjugate acid-base pair is $HSO_{4(aq)}^{-} - SO_{4(aq)}^{2-}$ .

8 0

3 years ago