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2 years ago

Which liquid materials have strong odor and weak odor?

1 answer:

5 0

Odor refers to the fragrance caused by one or more volatilized chemical compounds. It can be strong or weak. 
Strong odor have: Sodium Hypochlorite, Muriatic Acid, Sulphuric Acid, Ammunlom Sulfide. Butyl ,Butyric Acid, Pyridine

Weak Odors have Spray Glue, Dry Erase Markers, Paint cleaners. Water.

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Why does the Sun appear larger and brighter as seen from Earth than the other stars in the chart? Use the table to help you answ

FinnZ [79.3K]

Its C. for the first thing

8 0

2 years ago

Read 2 more answers

Identify and calculate the number of representative particles in 2.15 moles of gold.

LekaFEV [45]

Answer:

$1.29 * 10^{24}$ particles of gold

Explanation:

To convert the number of moles of any substance, in this case gold, you need Avogadro's number.

Avogadro's number is always $6.022$ × $10^{23}$

$2.15$ moles Au × $\frac{6.022*10^{23} particles}{1 mole Au}$ = $1.29 * 10^{24}$ particles of gold

5 0

2 years ago

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a

DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0

2 years ago

Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreas

-BARSIC- [3]

Answer:

The quantity of heat lost by the surroundings is 258,5J

Explanation:

The dissolution of salt XY is endothermic because the water temperature decreased.

The total heat consumed by the dissolution process is:

4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J

This heat is consumed by the calorimeter and by the surroundings.

The heat consumed by the calorimeter is:

42,2 J/°C × (0,93°C) = 39,2 J

That means that the quantity of heat lost by the surroundings is:

297,7J - 39,2J = 258,5 J

I hope it helps!

8 0

2 years ago

Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial rea

Dima020 [189]

Answer : The correct option is, (C) $^{58}\textrm{Fe}$

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

$_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma$

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e$

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

$_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma$

Process 2 : Beta emission.

$_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e$

Process 3 : Neutron capture.

$_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma$

From this we conclude that, the initial reactant in the production of cobalt-60 is $_{26}^{58}\textrm{Fe}$

Hence, the correct option is, (C) $^{58}\textrm{Fe}$

7 0

2 years ago