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vazorg [7]
3 years ago
14

Someone can help me ???

Chemistry
2 answers:
Oxana [17]3 years ago
7 0
The correct option is (a) Fruit salad.


As one cannot mix all the veges (or whatever you are using ) uniformly everywhere.

Tresset [83]3 years ago
6 0
C?? Think that is right
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The most negative electron affinity is most likely associated with which type of atoms?
Llana [10]

Explanation:

elements are based on electrical conductivity

7 0
3 years ago
Read 2 more answers
3. Sexual reproduction is the process of creating a new organism by combining the
weqwewe [10]

Answer:

chromosomes is responsible

6 0
2 years ago
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo s
Ray Of Light [21]

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

\frac{V}{T}=k

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

\frac{P}{T}=k

Combinado las mencionadas tres leyes se obtiene:

\frac{P*V}{T} =k

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

  • P1: 750 torr
  • V1: 8.5 L
  • T1: 20°C= 293°K (siendo 0°C=273°K)
  • P2: 425 torr
  • V2: ?
  • T2: -20°C= 253 °K

Reemplazando:

\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}

Resolviendo:

V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}

V2= 12.95 L

<u><em>El volumen del gas era 12.95 L</em></u>

<u><em></em></u>

5 0
2 years ago
PLEASE HELP!!!
trasher [3.6K]

Answer:

(a) oxygen

(b) 154g (to 3sf)

(c) 79.9% (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

4 0
2 years ago
You are given a solid that is a mixture of na2so4 and k2so4.
Murljashka [212]

Here we have to calculate the amount of SO_{4}^{2-} ion present in the sample.

In the sample solution 0.122g of SO_{4}^{2-} ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ +  SO_{4}^{2-}

(Na)₂SO₄=2Na⁺ +  SO_{4}^{2-}

Thus, BaCl₂+  SO_{4}^{2-} = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of  SO_{4}^{2-} ion is precipitated in this reaction.  

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion (SO_{4}^{2-}) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present. So in 1 g of BaSO₄ \frac{96.06}{233.3}=0.411 g of SO_{4}^{2-} ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

5 0
3 years ago
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