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Alenkinab [10]
3 years ago
8

What is uniform particles ?

Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
8 0

This particle is a co-polymer particle with a plastic base which can easily control arbitrary thickness. It also has excellent involtage, heat and chemical resistance.

hope this helps:)sorry if it doesnt

You might be interested in
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
Allisa [31]

Answer:

The mass of water m_{w} = 39.18 gm

Explanation:

Mass of iron m_{iron} = 32.5 gm

Initial temperature of iron T_{1} = 22.4°c = 295.4 K

Specific heat of iron  C_{iron} = 0.448 \frac{KJ}{kg K}

Mass of water = m_{w}

Specific heat of water  C_{w} = 4.2 \frac{KJ}{kg  K}

Initial temperature of water T_{2} = 336 K  

Final temperature after equilibrium T_{f} = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water  = Heat gain by iron rod

m_{w} C_{w} (T_{2} - T_{f} ) =  m_{iron} C_{iron} ( T_{f} - T_{1} )

Put all the values in above formula we get

m_{w} × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

m_{w} = 39.18 gm

Therefore the mass of water m_{w} = 39.18 gm

8 0
3 years ago
Why are fossils<br> important in the<br> development of the<br> geologic time scale?
olga55 [171]
Fossils are fundamental to the geologic time scale. The names of most of the eons and eras end in zoic, because these time intervals are often recognized on the basis of animal life. Rocks formed during the Proterozoic Eon may have fossils of relative simple organisms, such as bacteria, algae, and wormlike animals
4 0
2 years ago
I need help with this assignment
yuradex [85]

The answers for the following sums is given below.

1.Pd_{2}H_{2}

2.C_{2} H_{6}

3.C_{2} H_{2} O_{2} Cl_{2}

4.C_{3}Cl_{3} N_{3}

5.Tl_{2 } C_{4} H_{4}O_{6}

6.C_{8}H_{8}

7. N_{2}O_{5}

8.P_{4}O_{6}

9.C_{4}H_{8}  O_{2}

Explanation:

1.Given:

        Molar mass=216.8g

Molecular formula=PdH_{2}

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of PdH_{2}:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of PdH_{2}=106.4+2=108.4

n=\frac{216.8}{106.4}

<u><em>n=2</em></u>

Molecular formula=2(PdH_{2})

Molecular formula=Pd_{2}H_{2}

Therefore the molecular formula of the compound is Pd_{2}H_{2}

2. Given:

        Molar mass=30.0g

Molecular formula=CH_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of CH_{3}:

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of CH_{3}=12.01 + 3 =15.01u

n=\frac{30.0}{15.01}

<u><em>n=2</em></u>

Molecular formula=2(CH_{3})

Molecular formula=C_{2} H_{6}

Therefore the molecular formula of the compound is C_{2} H_{6}

3. Given:

        Molar mass=129g

Molecular formula=CHOCl

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n=\frac{129}{64.5}

<u><em>n=2</em></u>

Molecular formula=2(CHOCl)

Molecular formula=C_{2} H_{2} O_{2} Cl_{2}

Therefore the molecular formula of the compound is C_{2} H_{2} O_{2} Cl_{2}

5. Given:

        Molar mass=577g

Molecular formula=TlC_{2} H_{2}O_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of TlC_{2} H_{2}O_{3} :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of TlC_{2} H_{2}O_{3}=204.3+24.02+1+48.00=278.32u

n=\frac{577}{278.32}

<u><em>n=2</em></u>

Molecular formula=2 (TlC_{2} H_{2}O_{3})

Molecular formula=Tl_{2 } C_{4} H_{4}O_{6}

Therefore the molecular formula of the compound is Tl_{2 } C_{4} H_{4}O_{6}

4. Molar mass=184.5g

Molecular formula=CClN

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n=\frac{184.5}{61.5}

<u><em>n=3</em></u>

Molecular formula=3 (CClN)

Molecular formula=C_{3}Cl_{3} N_{3}

Therefore the molecular formula of the compound is C_{3}Cl_{3} N_{3}

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula=C_{8}H_{8}

Molecular formula =C_{8}H_{8}

Molar mass of C_{8}H_{8}:

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of C_{8}H_{8}=96.08+8=104.08u

n=104.08÷78.0

<em><u>n=1</u></em>

Molecular formula = n(Empirical formula)

Molecular formula = 1(C_{8}H_{8})

Molecular formula =C_{8}H_{8}

Therefore the molecular formula of a compound is C_{8}H_{8}

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen=\frac{4.04}{14.01} = 0.289 moles

moles of oxygen=\frac{11.46}{15.999} =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula  of N_{2} O{5}.

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is N_{2}O_{5}.

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula=P_{2} O_{3}

molecular formula= 2(Empirical formula)

Molecular formula =2(P_{2} O_{3})

Molecular formula =P_{4}O_{6}

Therefore the molecular formula of the compound is P_{4}O_{6}

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula =C_{2} H_{4} O

Molecular formula = 2(Empirical formula)

Molecular formula =2(C_{2} H_{4} O)

Molecular formula =C_{4}H_{8}  O_{2}

Therefore the molecular formula of the compound is C_{4}H_{8}  O_{2}

4 0
3 years ago
During the third stage of labor what happens? (it's not C)
QveST [7]

Answer:

B

Explanation:

placenta comes after baby

5 0
2 years ago
How many moles are in sample containing 2.71 x 10^24 atoms of iron?
Deffense [45]

Answer:

4.5 moles

Explanation:

One mole is equal to 6.022 x 10^23 atoms

2.71 x 10^24 atoms * 1 mol/ 6.022 x 10^23 atoms = 4.5 moles

5 0
3 years ago
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