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Gemiola [76]
3 years ago
8

Can you please help me with these true and false questions thanks!

Chemistry
1 answer:
jek_recluse [69]3 years ago
6 0
1 true
2 false 
3 false
4 true
5false
6 false
7false
8true
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What is the noble gas electron configuration of bismuth (Bi)? A. [Kr] 6s2 6p3 B. [Xe] 6s2 6p3 C. [Xe] 4f14 5d10 6s2 6p3 D. [Kr]
OlgaM077 [116]

The answer is C) Xe 4f14 5d10 6s2 6p3

4 0
3 years ago
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At the freezing point, the particles in an object have no kinetic energy. true or false?
Sliva [168]
False because particles stop moving or study slow down.
5 0
3 years ago
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Earth rotates west to east. Is this true or false? I want to say false, but I'm not sure.
777dan777 [17]
The answer is false because it goes in a full 360 degree circle 
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3 years ago
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A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperatu
lapo4ka [179]

Answer:

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

6 0
2 years ago
By how much will the ph change if 0.025 mol of hcl is added to 1.00 l of the buffer that contains 0.15 m hc2h3o2 and 0.25 m c2h3
Setler [38]
According to the reaction equation:

              CH3COO-  + H+   → CH3COOH 

initial         0.25                             0.15

change     - 0.025                          + 0.025

Equ         (0.25-0.025)                 (0.15 + 0.025)

first, we have to get moles acetate and moles acetic acid:

moles of acetate = 0.25 - 0.025 = 0.225 moles

∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M

moles of acetic acid = 0.15 + 0.025 = 0.175 moles

∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M 

Pka = -㏒ Ka 

       = -㏒ 1.8 x 10^-5
       
       = 4.74

from H-H equation we can get the PH value:

PH = Pka + ㏒ [acetate / acetic acid]

PH = 4.74 + ㏒[0.225/0.175]

∴ PH = 4.8
3 0
3 years ago
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