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gayaneshka [121]

3 years ago

6

Which of the following describes an endothermic reaction?

1 answer:

Alenkinab [10]3 years ago

4 0

The endothermic reaction would be D. Batter becomes a pancake when heated. This is because the batter is absorbing the heat in order to turn into a pancake.

Answer is D. Batter becomes a pancake when heated.

Please vote brainliest thank you!

You might be interested in

If 22.5L of nitrogen at 0.98 atm is compressed to 0.95 atm,what is the new volume?

Ksenya-84 [330]

At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

<h3>What is Boyle's law?</h3>

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Given that;

Initial volume of the gas V₁ = 22.5L

Initial pressure of the gas P₁ = 0.98atm

Final pressure of the gas P₂ = 0.95atm

Final volume of the gas V₂ = ?

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = (0.98atm × 22.5L) / 0.95atm

V₂ = 22.05Latm / 0.95atm

V₂ = 23.2L

Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.

Learn more about Boyle's law here: brainly.com/question/1437490

#SPJ1

6 0

2 years ago

You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately to 4 significant fi

zmey [24]

Answer:

Average density for method A = 2.4 g/cm³

Average density for method B = 2.605 g/cm³

Explanation:

In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):

For Method A:

Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6

Average = 9.6/4 = 2.4 g/cm³

For Method B:

Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420

Average = 10.420/4 = 2.605 g/cm³

5 0

3 years ago

When the pH decreases from 6 to 3, the hydrogen ion, H+, increases by what?

Irina-Kira [14]

pH decreases as the hydrogen ion concentration increases.

<u>Explanation:</u>

When there is a decrease in pH, that is pH decreases from 6 to 3 then the acidity increases.

That is the pH is between 1 to 7 then it is acidic

When the pH is 7 then it is neutral

When the pH is between 7 to 14 then it is basic

As the H⁺ ion concentration increases, then the pH value decreases, here pH decreases from 6 to 3.

So the concentration of Hydrogen ion increases, pH decreases.

3 0

3 years ago

Select all that apply. Using your periodic table of the elements which chemical symbols indicate an element?

tatuchka [14]

Answer:

O, I, Ca

Explanation:

4 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago