All categories

2 years ago

¿la densidad de una solución es mayor o menor al agregar agua?

Chemistry

1 answer:

andreev551 [17]2 years ago

5 0

Answer:

La densidad experimenta una disminución

Explanation:

Hola!

En este caso, debemos recordar que la adición de agua incrementa el volumen donde se encuentra un soluto disuelto, el cual posee una cantidad de materia constante. De este modo, sabiendo que la densidad es:

$\rho =\frac{m}{V}$

Al incrementar el denominador (volumen), la densidad experimenta una disminución, al estar en relación inversamente proporcional.

Muchos saludos!

You might be interested in

Can atoms compose molecules?

Nana76 [90]

Answer: yes

Explanation: Atoms can join together to form a molecule

7 0

3 years ago

Read 2 more answers

PLS HELP IF YOU WANT A BRAINLIEST BEST ANSWER CHOICE GETS BRAINLIEST Which describes the two cells formed through mitosis?

luda_lava [24]

Answer:C

Explanation:

did it

6 0

2 years ago

Helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

vlabodo [156]

Answer:

1. Delete lion, bison, reindeer, and giraffe

2. Delete cactus, oak, evergreen,

Explanation:

1. Lions live in the African savanna, bison live in open fields with lots of grass, reindeer live in Antarctica and other very cold places, and giraffes also live in the areas of the African savanna where there are trees.

2. Cacti live in the desert, oaks and evergreens live in open grassy areas.

Hope this helps

5 0

2 years ago

Read 2 more answers

ammonia gas reacts with oxygen gas, o2(g), to produce nitrogen dioxide and water. when 28.5 g of ammonia gas reacts with 83.4 g

jolli1 [7]

The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.

A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)

NH₃ + O₂ → NO₂ + H₂O

The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O

Calculate the number of moles from the reactant

Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol
Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol
n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.

According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.

Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol
The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol

The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.

P = (1.35 ÷ 1.49) × 100%

P = 0.9052 × 100%

P = 90.52%

Learn more about stoichiometry here: brainly.com/question/13691565

#SPJ4

7 0

1 year ago

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0

2 years ago