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3 years ago

EXPLAIN: How does the amount of a substance affect the rate at which temperature changes?

1 answer:

4 0

The more of a substance you have the longer the reaction will take. If you have a big pot of water and a little pot of water. The little pot will heat up quicker then the bigger pot because it does not need to fill a space up. Think of it as diffusion. When you have a 500 and 50 mL beaker, and you add a blue and red dye to them both. It will only take about a minute for the dye to almost completely mix together in the 50 mL. While the 500 mL it will take much longer for dye to mix all together then it did in the 50 mL. When it is a bigger substance it takes longer for a reaction to fully expose the substance to that reaction. If you have food, if you only have one piece of food you only need to put it in for 30 seconds to 1 minute. While if you have multiple pieces of food you will need longer time to be able to heat all of the pieces. If you have a small bowl of soup and you put it in for a minute the middle will be warm. But if you had a bigger bowl and still put it in for only a minute then the middle will be cold because you did not give it enough reaction time to heat the whole substance.

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If i initially have 4.0 l of a gas at a pressure of 1.1 atm, what will the volume be if i increase the pressure of 3.4 atm?

ddd [48]

The volume of a gas that its pressure increase to 3.4 atm is calculated as follows

By use of boyles law that is P1V1=P2V2
V1=4.0 L
P1=1.1 atm
P2=3.4 atm
V2= P1V1/P2

(1.1 atm x 4.0 L)/3.4 atm= 1.29 L

4 0

2 years ago

How do plants release energy

ololo11 [35]

Answer:

no but oxygen

Explanation:

no but oxygen

7 0

3 years ago

Read 2 more answers

Classify the following reactions: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l) Ba(OH)2(aq)+ZnCl2(aq)→BaCl2(a

Savatey [412]

The reactions are

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Ba(OH)2(aq) + ZnCl2(aq) → BaCl2(aq) + Zn(OH)2(s)

2AgNO3(aq) + Mg(s) → Mg(NO3)2(aq) + 2Ag(s)

Answer:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) = Acid-base reaction

Ba(OH)2(aq) + ZnCl2(aq) → BaCl2(aq) + Zn(OH)2(s) = Precipitation reaction

2AgNO3(aq) + Mg(s) → Mg(NO3)2(aq) + 2Ag(s) = Redox reaction

Explanation:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) .

The reaction above is an example of an acid-base reaction. Acid-base reaction is a chemical reaction between an acid and a base. The product is usually a salt and water. An acid dissolves in water to produce H+ ion. Hydrochloric acid is a strong acid as it is completely dissociated in water. A base dissolves in water to form hydroxide ion(OH-). The sodium hydroxide is a strong base as it completely dissociate in water. The reaction between a base and an acid can be known as a neutralization reaction.

Ba(OH)2(aq) + ZnCl2(aq) → BaCl2(aq) + Zn(OH)2(s)

The reaction above is a precipitation reaction. Precipitation reaction is a reaction where two ionic bonds or cations and anions in an aqueous solution react to form an insoluble salts. The insoluble salt formed is known as the precipitate. The reaction above is an example of a double replacement reaction. From the equation the ions replace each other depending on the cations and anions. They switch partners as both reactants lose their partners to form new partnership with a different ion. The two reactants ( Ba(OH)2 and ZnCl2 ) are aqueous solution, they react to from a solid precipitates( Zn(OH)2(s) ).

2AgNO3(aq) + Mg(s) → Mg(NO3)2(aq) + 2Ag(s)

The reaction is known as an oxidation-reduction reaction(Redox reaction). A oxidation-reduction reaction their is a transfer of electron(s) between two species. The oxidation number of a chemical reaction changes by losing or gaining electrons. One of the reactant is a reducing agent and the other is an oxidizing agent.

From the reaction Mg is the reducing agent and AgNO3 is the oxidizing agents.

2 Ag+ + 2 e- → 2 Ag∧0 (reduction)

Mg∧0 - 2 e- → Mg²+ (oxidation)

8 0

3 years ago

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 264 mL .

gavmur [86]

The formula for density is:

D = m/v

We can use the formula to figure out the mass because we already know two of the three values (we are given the density and volume), so we only have to solve for m. If we plug our given values into the formula, we get:

2.70 = m / 264

Now, all we need to do is solve for m. The goal is to get m on one side of the equation, and all we have to do is multiply each side of the equation by 264:

264 × 2.70 = (m÷264) × 264

264 × 2.70 = m

m = 712.8

The mass of the piece of aluminum is 712.8 grams.

4 0

3 years ago

Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

Solubility in 0.0120 M CoBr₂ (S')

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0

2 years ago