When baking soda and vinegar react, the product includes bubbles. what mostly likely occurred?

Describe the double circulation system of the human body

Alika [10]

Answer:

we have two loops in our body in which blood circulates. One is oxygenated, meaning oxygen rich, and the other is deoxygenated, which means it has little to no oxygen, but a lot of carbon dioxide.

4 0

3 years ago

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Alona [7]

Answer:

Approximately $1.95 \times 10^{2}\; \rm J$ .

Explanation:

Look up the specific heat of gaseous neon:

$c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}$ .

Calculate the required temperature change:

$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Let $m$ denote the mass of a sample of specific heat $C$ . Energy required to raise the temperature of this sample by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

For the neon gas in this question:

$c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}$ .
$m = 10.6\; \rm g$ .
$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Calculate the energy associated with this temperature change:

$\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}$ .

3 0

3 years ago

What are the units used to measure specific heat capacity?

Andrews [41]

The units used to measure specific heat capacity is Joules per kilogram per Kelvin.

<h3>What is specific heat capacity?</h3>

It is the amount of heat absorbed per kilogram of material when the temperature rises by 1 Kelvin.

Specific heat capacity C is the Joules of energy in form of heat per kilogram per Kelvin temperature. The units represented by

C = ___ J/kg.K

Thus, the units used to measure specific heat capacity is Joules per kilogram per Kelvin.

Learn more about specific heat capacity.

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7 0

2 years ago

Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)

Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g) K₁= ? (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g) K₂= 3.61 (2)

2 A(g) + D(g) ⇄ C(g) K₃= 7.19 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3) algebraically to get our desired equilibrium (1).

We are allowed to reverse reactions, in that case we take the reciprocal of K as our new K' ; we can also add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3) and we want D as a reactant . That suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B as product in our equilibrium (1) . Finally we would add (2) and (3) to get (1) which is our final goal.

2C(g) ⇄ 2A(g) + 2B(g) K₂´= ( 1/ 3.61 )²

₊

2 A(g) + D(g) ⇄ C(g) K₃ = 7.19

C(g) + D(g) ⇄ 2B(g) K₁ = ( 1/ 3.61 )² x 7.19

K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves gases.

To compute the last part lets setup the following mnemonic ICE table to determine the quantities at equilibrium:

pressure (atm) C D B

initial 1.64 1.64 0

change -x -x +2x

equilibrium 1.64-x 1.64- 2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)² where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0

3 years ago

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

erastovalidia [21]

Answer:

The time taken for the cross mark to disappear decreases steadily down the column.

Explanation:

Now if we look at the data provided, we will discover that the volume of the HCl was held constant while the volume of the thiosulphate was increased steadily and the volume of water decreased steadily.

Recall that a system is more concentrated when it contains less volume of water and more volume of reactants. Hence as the volume of water in the system is being reduced, the concentration of reactants is increased.

It has been established that an increase in the concentration of reactants lead to an increase in the rate of reaction. The disappearance of the cross shows the completion of the reaction between HCl and thiosulphate. The faster or slower the cross disappears, the faster or slower the rate of reaction.

Since increase in concentration of reactants increases the rate of reaction, it is observed that as the volume of the thiosulphate increases (reactant concentration increases) the cross disappears faster (rate of reactant increases). Hence as the volume of thiosulphate increases, it takes a shorter time for the cross to disappear. This implies that the time column in the table (refer to the question) will decrease steadily as the volume of thiosulphate increases.

5 0

3 years ago