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BabaBlast
4 hours ago
8

per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 400.m3. P is destro

yed by sunlight, and once in the pond it has a half-life of 46.h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits.
Chemistry
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A piece of lead loses 78.0 J ofheat and experiences a decrease in temperature of 9.0C the specific heat of lead is .130J/gC what
KiRa [2933]
To determine the mass of the lead piece, we use the following equation: Q(heat) = mC delta T, where Q equals 78.0 j, M is the mass we want to find, C is the specific heat capacity (0.130 j/g/C), and delta T shows the temperature difference, set at 9.0 c. Rearranging the formula to solve for M gives us M = Q / c delta T. By substituting in the values, we conclude that M = 78.0 j / (0.130 j/g/C * 9.0 C), calculating this gives us a mass of 66.7 g of lead.
3 0
2 months ago
A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
KiRa [2933]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

Por lo tanto,

0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

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An electric kettle uses electrical energy to boil water. Energy from the electricity is transferred to the water, heating it up.
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In the electrical ice maker, water releases energy as it freezes into ice, translating this energy into bond energy. Although there are hydrogen bonds in the liquid water state, adding electrical energy converts the water from liquid to solid, increasing bond strength. The potential energy from the water is now represented as hydrogen bond energy in the ice.
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Pyridine rings can also under electrophilic aromatic substitution. given 2-methoxypyridine below, draw the expected major produc
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Co2 is indeed the correct answer, my friend.
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