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Fudgin [204]
2 years ago
10

In Millikan’s experiment, the oil droplets acquire one or more negative charges by combining with the negative charges that are

produced from the ionization of air by X rays. By measuring the charges on the oil droplets, he calculated the charge on a single electron as −1.60×10−19 C. The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron.If Millikan was measuring the charge on an oil droplet with 4 negatively charged electrons on it, what charge would he have measured on the droplet?
Chemistry
1 answer:
zimovet [89]2 years ago
4 0

Answer:

-6.4x10⁻¹⁹ C

Explanation:

The elementary charge of one electron is -1.60x10⁻¹⁹C, so each electron has its charge, and a sample with more than one electrons will have a multiple of its charge, which is proportional to the number of electrons. So, if the oil droplet had 4 electrons, thus the charge will be four times the elementary charge:

4*(-1.60x10⁻¹⁹) C = -6.4x10⁻¹⁹ C

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Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0
2 years ago
Predict all bond angles in the following compound. Be sure to answer all parts. HCa≡CbCcH2OH
morpeh [17]

Answer:

See explanation below

Explanation:

What we have to consider is the hybridation of the three carbon atoms we are asked  in this question .

Hybridization   # bonds                            Angle

sp³                         4                                   109.5º

sp²                         3 +   1 pi bond              120º

sp                           2 + 2 pi bonds             180º

Carbon atom (a) is bonded to two atoms: Carbon (b) and one Hydrogen. It has a triple bond to Carbon (b). Therefore its hybridization is sp with two pi bonds, and for sp hybridization we know the angle is 180 º.

The same hybridization sp happens to carbon (b) bonded to Carbon (c) and C(a) using one sp bond to Carbon (a) and 2 pi bonds; it is also bonded using the other sp  to Carbon (c). The angle is therefore 180 between Carbons b and c.

Carbon C is bonded to 4 atoms, therefore, its hybridization is sp³ and the angles with these 4 atoms will be 109.5 º tehedral  ( one bond to OH, one to C(b), and 2 to H ) .

6 0
2 years ago
A length of glass tubing is 0.525 m. How many inches long is the tubing? (2.54 cm = 1 inch)
lara31 [8.8K]
If, (.525m)(100 cm/m)(1/2.54 in/cm) = 20.7 in  or C
6 0
2 years ago
Read 2 more answers
The melting point of water is 0°C at 1 atm pressure because under these conditions:
Tems11 [23]

Answer:

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:  

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:  

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.  

Now the value of entropy for surrounding ΔSsurr will be,  

ΔSsurr = -ΔHfusion/T  

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.  

8 0
2 years ago
vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu
Kaylis [27]

Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

5 0
2 years ago
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