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Elis [28]
3 years ago
13

Another word for "basic" is?

Chemistry
2 answers:
irakobra [83]3 years ago
8 0
Simple...I looked this up on google..hope this helps
natita [175]3 years ago
5 0

Answer: Fundamental, Key, vital, crucial

Explanation:

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Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?
monitta

Answer:

200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm (the second choice) would contain an equal number of gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\! (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, 200\; \rm mL of neon \rm Ne\, (g) at 25^{\circ} \rm C and 1\; \rm atm, is the only choice where the volume of the sample is also 200\; \rm mL \!. Hence, that choice would be the only one with as many gas particles as 200\; \rm mL\! of \rm He \, (g) at 25^{\circ} \rm C\! and 1\; \rm atm\!.

7 0
3 years ago
Each element is____than____element.
Ahat [919]
Each element is diffrent than another element 
6 0
3 years ago
Most marble is composed primarily of
KatRina [158]

Answer:

It is composed primarily of the mineral calcite (CaCO3) and usually contains other minerals, such as clay minerals, micas, quartz, pyrite, iron oxides, and graphite.

Explanation:

Marble can be otherwise called metamorphic rock which is formed as a result of heat application on limestone and pressure of metamorphism. It is composed primarily of the mineral calcite (CaCO3) and usually contains other minerals, such as clay minerals, micas, quartz, pyrite, iron oxides, and graphite.

8 0
4 years ago
Read 2 more answers
Complete the electron-dot structure of s-allylcysteine, showing all lone-pair electrons.
erastova [34]
The structural formula of <span>s-allylcysteine is shown in the picture (top figure). To create its Lewis structure, draw all its bonds between elements. Each single bond contains two electrons. There is an octet rule that must be obeyed by most elements. Each element should be surrounded with 8 electrons. The hydrogen is exempted of this rule. So, there are 4 lone pairs for the S atom, 1 lone pair for the N atom, and 2 lone pairs each for the 2 O atoms.</span>

7 0
3 years ago
Please need this ASAP. Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone,
stiv31 [10]

Answer:

1.4×10⁸ g of CaO

Explanation:

We'll begin by converting 250 tonnes to grams (g). This can be obtained as follow:

1 tonne = 1×10⁶ g

Therefore,

250 tonne = 250 × 1×10⁶

250 tonne = 2.5×10⁸ g

Next, the balanced equation for the reaction.

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Finally, we shall determine the mass of CaO produced by the decomposition of 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃. This can be obtained as follow:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 2.5×10⁸ g of CaCO₃ will decompose to produce =

(2.5×10⁸ × 56)/100 = 1.4×10⁸ g of CaO.

Thus, 1.4×10⁸ g of CaO will be obtained from 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃.

6 0
3 years ago
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