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3 years ago

The repeating pattern of a mineral’s particles forms a solid called a(n)

Chemistry

2 answers:

natali 33 [55]3 years ago

8 0

Answer:

The answer is a

Explanation:

bazaltina [42]3 years ago

6 0

Crystalline solids are solids that have particles arranged in a regular, repeating patterns. They are held together by uniform intermolecular forces. The faces intersect at a specific angle that identifies that substance. They have distinctive internal structures that in turn lead to distinctive flat surfaces. These flat surfaces give a short rage order. When exposed to x – ray diffractometers, they exhibit a distinctive pattern used to identify the material. They give the same face when you turn a crystalline solid around. The answer is crystal.

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What mass of water is needed to react with 32.9 g Li3N?

laiz [17]

Answer:

The answer to your question is 50.76 g of H₂O

Explanation:

Data

mass of water = ?

mass of Li₃N = 32.9 g

Process

1.- Write the balanced chemical reaction

Li₃N + 3H₂O ⇒ 3LiOH + NH₃

2.- Calculate the molar mass of Li₃N and H₂O

Li₃N = (3 x 7) + (1 x 14) = 35 g

H₂O = (2 x 1) + (16 x 1) = 18 g

3.- Use proportions to calculate the mass of water

35 g of Li₃N ---------------- (3 x 18) g of water

32.9 g of Li₃N ------------ x

x = (32.9 x 54)/35

x = 1776.6/35

x = 50.76 g of H₂O

8 0

3 years ago

Show how to calculate the mass of a proton, neutron, and electron in kilograms using the data and the equality statement 1 amu =

barxatty [35]

Answer:

Proton: m= 1.6726x10⁻²⁷ kg

Neutron: m= 1.6749x10⁻²⁷ kg

Electron: m= 9.1164x10⁻³¹ kg

Explanation:

We can calculate the mass of a proton, neutron, and electron using the following data:

mass of proton: 1.00728 amu

mass of neutron: 1.00867 amu

mass of electron: 5.49x10⁻⁴ amu

1 amu = 1.66054x10⁻²⁷ kg

Now, the mass of a proton, neutron, and electron in kilograms can be calculated using the next relation:

$m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot particle's mass (amu)$

For the proton:

$m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 1.00728 amu = 1.6726 \cdot 10^{-27} kg$

For the neutron:

$m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 1.00867 amu = 1.6749 \cdot 10^{-27} kg$

For the electron:

$m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 5.49x10⁻⁴ amu = 9.1164 \cdot 10^{-31} kg$

I hope it helps you!

4 0

3 years ago

All of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxyg

tankabanditka [31]

Answer:

High partial pressure of oxygen in the tissues.

Explanation:

Hemoglobin is a protein found in the red blood cells that is responsible for the transportation of oxygen from the lungs to the rest of the tissues in the body. The partial pressure determines whether oxygen is loaded or unloaded into the hemoglobin.

When there is lack of oxygen in our bodies, we experience hypoxia.

6 0

3 years ago

Flammability is a material’s ability to burn in the presence of

frosja888 [35]

Answer:

Flammability is a material’s ability to burn in the presence of oxygen.

Explanation:

Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.

The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.

5 0

3 years ago

1125 J of energy is used to heat 250 g of iron to 55 °C. The specific heat capacity of iron is 0.45 J/(g·°C).

Stolb23 [73]

Answer:

45 °C.

Explanation:

From the question given above, the following data were obtained:

Heat (Q) = 1125 J

Mass (M) = 250 g

Final temperature (T₂) = 55 °C

Specific heat capacity (C) = 0.45 J/gºC

Initial temperature (T₁) =?

The initial temperature of the iron can be obtained as illustrated below:

Q = MC(T₂ – T₁)

1125 = 250 × 0.45 (55 – T₁)

1125 = 112.5 (55 – T₁)

Divide both side by 112.5

1125/112.5 = 55 – T₁

10 = 55 – T₁

Collect like terms

10 – 55 = –T₁

–45 = –T₁

Multiply through by –1

45 = T₁

T₁ = 45 °C

Therefore, the initial temperature of the iron is 45 °C

8 0

3 years ago