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3 years ago

What type of minerals are these??

Chemistry

2 answers:

skelet666 [1.2K]3 years ago

6 0

Answer:

talc mineral and magnetite

Explanation:

Margaret [11]3 years ago

3 0

It’s gold and salt !

if it’s write mark as brainiest?

If it’s wrong please comment how can i know which type minerals is it?

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a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor

Diano4ka-milaya [45]

<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

= (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

= (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

X = 32.978°C

= 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0

4 years ago

A chemical processing plant has a network of pipes that are used to transfer liquid chemicalproducts from one part of the plant

liq [111]

Answer:

Explanationis the long-distance transportation of a liquid or gas through a system of pipes—a pipeline—typically to a market area for consumption. The latest data from 2014 gives a total of slightly less than 2,175,000 miles (3,500,000 km) of pipeline in 120 countries of the world.[1] The United States had 65%, Russia had 8%, and Canada had 3%, thus 75% of all pipeline were in these three countries.[1]

Pipeline and Gas Journal's worldwide survey figures indicate that 118,623 miles (190,905 km) of pipelines are planned and under construction. Of these, 88,976 miles (143,193 km) represent projects in the planning and design phase; 29,647 miles (47,712 km) reflect pipelines in various stages of construction. Liquids and gases are transported in pipelines and any chemically stable substance can be sent through a pipeline.[2] Pipelines exist for the transport of crude and refined petroleum, fuels – such as oil, natural gas and biofuels – and other fluids including sewage, slurry, water, beer, hot water or steam for shorter distances. Pipelines are useful for transporting water for drinking or irrigation over long distances when it needs to move over hills, or where canals or channels are poor choices due to considerations of evaporation, pollution, or environmental impact.:

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3 years ago

What is the wavelength, in nm, of the light photon emitted by a hydrogen atom when an electron goes from n

xxMikexx [17]

Answer:

Hence, the wavelength of the photon associated is 1282 nm.

Explanation:

8 0

3 years ago

Rank the following atoms by number of valence electrons. Rank from most to fewest valence electrons. To rank items as equivalent

Fed [463]

To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.

Rb: [Kr] 5s¹ - 1 valence electron

Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons

Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons

I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons

In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb

6 0

3 years ago

Read 2 more answers

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0

3 years ago