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Katen [24]
3 years ago
5

PLSSS HELP ME THIS IS FOR SIENCE PLLLLSSSSS

Chemistry
2 answers:
pickupchik [31]3 years ago
8 0

Answer:

Conduction

Explanation:

I took the course

Natali [406]3 years ago
8 0

Answer:

B. conduction

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What kind of matter is smoke?
Neporo4naja [7]

Answer : Colloid

Explanation : Any colloid which consist of a solid evenly dispersed in a gas phase can be called as smoke. So, if smoke has to be given a matter of substance then it is can be categorized as a Colloid present in gaseous phase. Smoke is a solid which contains minute particles of dust which is homogeneously dispersed in the air.

5 0
3 years ago
Read 2 more answers
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
14.A sample of fluorine gas has a density of _____
yarga [219]

Answer:

d = 0.793 g/L

Explanation:

Given data:

Density of fluorine gas = ?

Pressure of gas = 0.554 atm

Temperature of gas = 50 °C (50+273.15K = 323.15 K)

Solution:

Formula:

PM = dRT

M = molar mass of gas

P = pressure

R = general gas constant

T = temperature

d = PM/RT

d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K

d = 21.05 atm.g/mol/26.53 atm.L /mol

d = 0.793 g/L

8 0
3 years ago
The blank is the accepted form of gathering and reporting information within the science community.
Charra [1.4K]

Answer:

Scientific method.

Explanation:

Scientific method is the way taken to acquire scientific knowledge. It includes experiments, statistical analysis of existing data, and all kinds of observations of the world around us, while theoretical research is based on deriving certain theories about the world from basic principles, in a mathematical or logical way. The scientific method applies to both types of research, and emphasizes that scientific research is objective, that it can be verified by other scientists, and that knowledge is not acquired without context, but in a way that leads to a greater understanding of previous research and the world. we live in. To contribute to this, researchers are expected to clearly record both their findings and the methods they use to arrive at the results.

8 0
3 years ago
What are the symbols for the elements with the following valence electron configurations? List all.a. s2d1b. s2p3c. s2p
maria [59]

Answer:

Sc : 4s 2 3d 1

 Y : l 5s 2 4d 1

La : 6s 2 5d 1

Ce : 6s 2 4f 1 5d 1

Gd : 6s 2 4f 7 5d 1

Lu : 6s 2 4f 14 5d 1

Ac : 7s 2 6d 1

Pa : 7s 2 5f 2 6d 1

U : l 7s 2 5f 3 6d 1

Np : 7s 2 5f 4 6d 1

Cm : 7s 2 5f 7 6d 1

b. s2p3

he pnictogens:

N l : 2s 2 2p 3

P l : 3s 2 3p 3

As : 4s 2 3d 10 4p 3

Sb : 5s 2 4d 10 5p 3

Bi : 6s 2 4f 14 5d 10 6p 3

Mc : 7s 2 5f 14 6d 10 7p 3

c.The noble gases:

Ne : 2s 2 2p 6

Ar : 3s 2 3p 6

Kr : 4s 2 3d 10 4p 6

Xe : 5s 2 4d 10 5p 6

Rn : 6s 2 4f 14 5d 10 6p 6

Og : 7s 2 5f 14 6d 10 7p 6

Explanation: From the periodic tables we can drive elements with the electronic configuration

Sc : 4s 2 3d 1

 Y : l 5s 2 4d 1

La : 6s 2 5d 1

Ce : 6s 2 4f 1 5d 1

Gd : 6s 2 4f 7 5d 1

Lu : 6s 2 4f 14 5d 1

Ac : 7s 2 6d 1

Pa : 7s 2 5f 2 6d 1

U : l 7s 2 5f 3 6d 1

Np : 7s 2 5f 4 6d 1

Cm : 7s 2 5f 7 6d 1

b. s2p3

he pnictogens:

N l : 2s 2 2p 3

P l : 3s 2 3p 3

As : 4s 2 3d 10 4p 3

Sb : 5s 2 4d 10 5p 3

Bi : 6s 2 4f 14 5d 10 6p 3

Mc : 7s 2 5f 14 6d 10 7p 3

c.The noble gases:

Ne : 2s 2 2p 6

Ar : 3s 2 3p 6

Kr : 4s 2 3d 10 4p 6

Xe : 5s 2 4d 10 5p 6

Rn : 6s 2 4f 14 5d 10 6p 6

Og : 7s 2 5f 14 6d 10 7p 6

3 0
3 years ago
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