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3 years ago

What can be made in a laboratory

2 answers:

8 0

Synthetic gem crystals have been manufactured since the late 1800s, and their production is often marked by a need for them in industrial applications outside of the jewelry industry. The first success was in producing synthetic ruby of faceting quality. Synthetic crystals are used in communications and laser technology, microelectronics, and abrasives. Because synthetics for jewelry applications can be “made to order” [i.e. consistent color and crystal shape] given the right ingredients, time, and the facilities to grow them, they are likely to be much less rare than natural gems of equal size, clarity, and saturation of color. Because of this, and because it is possible to confuse them with gems that are naturally occurring, there are strict guidelines regarding how they are marketed and sold. so that is you answer

ss7ja [257]3 years ago

8 0

I'm not really sure how to answer this. Like, what can't be made in a laboratory?

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The elements of the group IA are termed as alkali metals, because their ___ are alkaline.

Vadim26 [7]

The answer is hydroxides.

The elements of the group IA are termed as alkali metals, because their hydroxides are alkaline.

7 0

1 year ago

Why do we use scientific notation or scientific exponential numbers in math calculations?

seraphim [82]

I believe the answer is: in order not to write very big or very small number values

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3 years ago

In two or more complete sentences, explain how to balance the combustion equation: ___C5H12 + ___O2 ⟶ ___CO2 + ___H2O

egoroff_w [7]

Answer:

First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.

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3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

3 years ago

When 0.875 g. of magnesium was heated in air, 1.25 g. of magnesium oxide

Ilia_Sergeevich [38]

Answer:

1. What is the coefficient for calcium in the balanced equation?

2 Ca + 5 NaOH → 4 Na + 2 Ca(OH)2

2. The burning of 20.0 g of acetylene (C2H2) produces what mass of carbon dioxide? You must

write the equation, balance it and solve the mass-mass problem.

a. Balanced equation:

2 C2H2 + 5 O2  4 CO2 + 2 H2O

Explanation:

3 0

3 years ago