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expeople1 [14]
3 years ago
5

1. Which one is the odd one out? 0 CaO o Cu o CO o CH

Chemistry
1 answer:
Sauron [17]3 years ago
4 0

Answer:

Yttrium barium copper oxide (YBCO) is a family of crystalline chemical compounds, famous for displaying high-temperature superconductivity.It includes the first material ever discovered to become superconducting above the boiling point of liquid nitrogen (77 K) at about 92 K.Many YBCO compounds have the general formula Y Ba 2 Cu 3 O 7−x (also known as Y123),

Explanation:

i hope this makes sense and helps man

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Once a slide is prepared and placed onto the microscope, the magnification and focus need to be altered. Describe how to change
Harlamova29_29 [7]

Answer:

我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈

Explanation:

我實際上不知道答案,我只是為了點我實際上不知道答案,我只是為了點數而這樣做,哈哈,祝你好運哈哈數而這樣做,哈哈,祝你好運哈哈

5 0
2 years ago
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If an atom has 5 protons, how many electrons does it have?
IRISSAK [1]

Answer:

5

Explanation:

An atom has the same number of protons and electrons

8 0
3 years ago
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5 0
3 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
The temperature of a 100.0 g sample of water is raised from 30degrees celsius to 100.0 degrees celsius. How much energy is requi
worty [1.4K]

Answer:

29260J

Explanation:

Given parameters:

Mass of water sample  = 100g

Initial temperature = 30°C

Final temperature  = 100°C

Unknown:

Energy required for the temperature change = ?

Solution:

The amount of heat required for this temperature change can be derived from the expression below;

     H  = m c (ΔT)

H is the amount of heat energy

m is the mass

c is the specific heat capacity of water  = 4.18J/g°C

ΔT is the change in temperature

Now insert the parameters and solve;

          H  = 100 x 4.18 x (100 - 30)

          H  = 100 x 4.18 x 70 = 29260J

6 0
2 years ago
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