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Genrish500
9 days ago
6

What is the mass of 0.75 moles of (NH4)3PO4?

Chemistry
2 answers:
Anarel [852]9 days ago
8 0

Answer:

The formula becomes N3H12PO4.

Explanation:

Tems11 [846]9 days ago
3 0
Your answer would be (NH4)3PO4.
You might be interested in
What is the specific heat of an unknown metal if 1.67 kcal of energy are required to raise the temperature of 79.2 g sample of t
VMariaS [1037]

Answer:

The solution to your inquiry is C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

Explanation:

Data

Q = 1.67 kcal

mass = 79.2 g

ΔT = 63.3°C

Formula

Q = mCΔT

Solving for C

C = Q/mΔT

Substituting values

C = 1.67/(79.2 x 63.3)

Simplifying

C = 1.67 / 5013.4

Final Result

C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

4 0
6 days ago
Read 2 more answers
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
Anarel [852]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density is defined as mass per unit volume, meaning volume can be calculated as:

V=\frac{m}{d}

By substituting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Consequently, the volume gained by the individual will be 5484.61 cm^{3}.

6 0
8 days ago
To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if
castortr0y [923]

Answer:

The correct options include choice 2, 3, and 6.

Explanation:

Density is identified as the mass of a substance per unit volume occupied by that substance.

Density=\frac{Mass}{Volume}

The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.

2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.

6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.

The same metals in both instances will yield consistent densities due to the fixed density of the metal.

7 0
3 days ago
Equimolar samples of CH3OH(l) and C2H5OH(l) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is at
Alekssandra [968]

Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.

Explanation:

To clarify the answer provided, let’s begin by defining some concepts.

The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.

3 0
15 days 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 [971]

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

3 0
14 days ago
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