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lapo4ka [179]
1 year ago
7

. Un cuerpo irregular se introduce dentro de una probeta que inicialmente tiene un volumen de 20 ml como muestra la figura Despu

és de la experiencia podemos afirmar que * El volumen del cuerpo es 3 ml. El volumen del cuerpo es cero. El volumen del cuerpo es igual al iniciar la experiencia. El volumen del cuerpo es 24 ml.
Chemistry
1 answer:
MrMuchimi1 year ago
6 0

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

El volumen del cuerpo es el mismo al principio porque el volumen no cambia si la temperatura permanece igual. Si cambiamos la temperatura i. mi. Al aumentar la temperatura, las moléculas comienzan a expandirse y se produce un aumento de volumen mientras que cuando disminuimos la temperatura, las moléculas de esa sustancia comienzan a contraerse y el volumen de esa sustancia disminuye. Entonces concluimos que el volumen depende de la temperatura.

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In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form
iris [78.8K]

Answer:

I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.

Explanation:

<em>I verify this is correct. </em>

6 0
1 year ago
50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide s
miskamm [114]

Answer:

49 g/L is the concentration of the acid

Explanation:

Firstly, we proceed to write the equation of reaction.

2NaOH + H2SO4 ——-> Na2SO4 + 2H2O

We can see that 1 mole of the base reacted with two moles of the acid.

kindly note that dm^3 is same as liter

Firstly, we need to get the concentration of the reacted sulphuric acid in g/L

we use the simple titration equation below;

CaVa/CbVb = Na/Nb

From the question;

Ca = ?

Va = 25 cm^3

Cb = 20 g/L

we convert this to concentration in mol/L

Mathematically, that is concentration in g/L divided by molar mass in g/mole

molar mass of NaOH = 40 g/mol

so we have; 20g/L / 40 = 0.5 mol/L

Vb = 50 cm^3

Na = 1

Nb = 2

Where C represents concentrations, V volumes and N , number of moles

Now, substitute the values;

Ca * 25/0.5 * 50 = 1/2

25Ca/25 = 0.5

So Ca = 0.5 mol/L

Now to get the concentration of H2SO4 in g/L

What we do is to multiply the concentration in mol/L by molar mass in g/mol

That would be 0.5 * 98 = 49 g/L

4 0
2 years ago
MUSCLES IN A KANGAROO'S LEGS WORK BECAUSE OF THE CELLS THAT MAKE UP THE MUSCLE. WHICH COMPONENT OF CELL THEORY DOES THIS BEST IL
Fed [463]

Answer:

b

Explanation:

4 0
1 year ago
Read 2 more answers
Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,
8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0
1 year ago
Molecular bromine is 24 percent dissociated at 1600 k and 1.00 bar in the equilibrium br2 (
dlinn [17]
Br2 == 2Br

24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2

=> partial pressure of Br, P Br = 0.24 bar, and
     partical pressure of Br2, P Br2 = 0.76 bar

kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758


3 0
1 year ago
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