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3 years ago

What appears to happen to the moon as it goes through the cucle

Chemistry

2 answers:

Aloiza [94]3 years ago

6 0

The way the earth Orbit

IRISSAK [1]3 years ago

3 0

What appears to happen is the moon disapears little by little. but what you actually see is the shadow of the earth blocking some light from the sun.

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Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

7 0

3 years ago

What are 4 general properties of matter?

Law Incorporation [45]

1. Solid
2. Liquid
3. Gas
4. Plasma

3 0

3 years ago

Read 2 more answers

Which of the following is the least reactive nonmetal? A. I B. F C. Br D. Cl

tatyana61 [14]

the answer is A. I (iodine)

no worries

5 0

3 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

galina1969 [7]

Answer:

$\%\, yield\, \, SO_3=82.29$

Explanation:

First write the balance eqation of chemical reaction:

$2S(s) +3O_2(g) \rightarrow 2SO_3(g)$

Remember writing any chemical reaction from its elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen $=5/32mol=0.16mol$

mass of sulpher given=6gram

mole of sulpher $=6/32mol=0.19mol$

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with $1.5\times 0.19$ i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole $SO_3$

1 mole of sulpher will give 2/3 mole $SO_3$

0.16 of sulpher will give 0.12 mole $SO_3$

mass of $SO_3$ =9.6gram this is theoreical production of $SO_3$

and

actual production of $SO_3$ =7.9gram

$\%\, yield \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100$

$\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29$

$\%\, yield\, \, SO_3=82.29$

3 0

3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

taurus [48]

Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

6 0

3 years ago