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Natalija [7]
3 years ago
12

(20 Points) Which definition of "coarse" matches its use in the phrase "coarse focus?"

Chemistry
2 answers:
vlada-n [284]3 years ago
7 0

Answer by YourHope:


Hi! :)


Which definition of "coarse" matches its use in the phrase "coarse focus?"


A) "rough in texture"


Have a BEAUTIFUL day~

Finger [1]3 years ago
6 0

C) crude or unrefined


The course focus is the rough or basic focus knob on a microscope.

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Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3)
nadya68 [22]

Answer:

CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:

The decomposition reaction always make two compounds from one.

The products always have simpler chemical structure, originated from a determined compound. This can happens  spontaneously or by a third party.

A notable example of decomposition is hydrolysis. As for example the case of water, which decomposes and generates oxygen and hydrogen gas

2H₂O (l) → 2 H₂ (g) + O₂ (g)

In this case, the calium carbonate decomposes into CaO and CO₂

These two, are the products of the decomposition.

Of course, the unique reactant is the Calcium Carbonate

The balanced equation is:

CaCO₃(s) → CaO(s) + CO₂(g)

6 0
3 years ago
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Which feature does an iron metal have?
notka56 [123]
A feature that an iron metal has is a sea of electrons
3 0
3 years ago
According to datons atomic theory
grin007 [14]

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4 0
3 years ago
Can someone please help me?
Olin [163]

Answer:

c.)the reactants have the same mass as the products

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hope it helps!

8 0
3 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
3 years ago
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