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lozanna [386]
3 years ago
14

How many grams in a kilogram,how many quarts are in a pint and if how many ml of h20 = 1g

Chemistry
2 answers:
vodomira [7]3 years ago
5 0

Answer:

1 g  +  10 g

Explanation:

spayn [35]3 years ago
4 0

Answer:

there are a thousand grams in a kilo

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The flotation process used in metallurgy involves Multiple Choice the roasting of sulfides. separation of gangue from ore. elect
densk [106]

The flotation process used in metallurgy involves the separation of gangue from ore.

<h3>How does the flotation process work?</h3>

The various wettability qualities of a material's surface are the foundation of flotation operations. The basic principles of flotation are quite similar to those of a sink and float process, where the materials' relative densities to the medium in which they are deposited determine the basis of the separation.

<h3>What is the process of separating minerals from gangue known as?</h3>

Mineral processing, mineral dressing, or ore dressing are all terms for the process of separating minerals from gangue. It is an important and frequently necessary part of mining. Depending on the type of minerals used, the process may be difficult.

Learn more about minerals here:-

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8 0
2 years ago
HELP ASAP ILL GIVE BRAINLIEST!!
Dahasolnce [82]

Of course we would experience them.

3 0
2 years ago
Which index fossil is most likely to be found in<br> the Allegheny Plateau?
Usimov [2.4K]

Answer:

amphibians

Explanation:

please mark this answer as brainliest

5 0
3 years ago
How much heat is absorbed when a 298.3 g piece of brass goes from 30.0 to 150
igor_vitrenko [27]

Answer:

Q = 1360.248 j

Explanation:

Given data:

Mass of brass = 298.3 g

Initial temperature = 30.0°C

Final temperature = 150°C

Specific heat capacity of brass = 0.038 J/g.°C

Heat absorbed = ?

SOLUTION:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30.0°C  

ΔT = 120°C

Q = 298.3 g × 0.038 J/g.°C × 120°C

Q = 1360.248 j

3 0
3 years ago
Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2
HACTEHA [7]

Answer:

The answer to your question is:  0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

                   1 mol of NO2 ---------------------  6.023 x 10 ²³ molecules

                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

                   x = 0.00061 mol

         

                 1 mol of NO2 ---------------------  46 kg of NO2

                 0.00061 mol     ------------------    x

                 x = 0.00061 x 46/1

                x = 0.028 kg of NO2

7 0
3 years ago
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