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pashok25 [27]
2 years ago
9

Cobalt has an atomic number of 27 and atomic mass of 59. How many neutrons does cobalt have? 

Chemistry
1 answer:
elena-s [515]2 years ago
3 0

Answer:

32

Explanation:

59-27

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30 mL of 0.6 M HCl solution is neutralized with 90 mL NaOH solution. What is the concentration of the base? Show all work Please
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From

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<em>V</em><em>b</em><em>=</em><em>9</em><em>0</em><em>m</em><em>l</em><em>s</em>

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<em>M</em><em>b</em><em>=</em><em>?</em>

<em>f</em><em>r</em><em>o</em><em>m</em>

<em>Mb</em><em>=</em><em><u>M</u></em><em><u>a</u></em><em><u>×</u></em><em><u>V</u></em><em><u>a</u></em><em><u>×</u></em><em><u>n</u></em><em><u>b</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>V</em><em>b</em><em>×</em><em>n</em><em>a</em>

<em>M</em><em>b</em><em>=</em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>×</u></em><em><u>3</u></em><em><u>0</u></em><em><u>×</u></em><em><u>1</u></em>

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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>9</em><em>0</em>

<em>M</em><em>b</em><em>=</em><em>0</em><em>.</em><em>2</em><em>M</em>

<em>NaOH</em><em>=</em><em>4</em><em>0</em>

<em>F</em><em>r</em><em>o</em><em>m</em>

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3 0
2 years ago
In a laboratory experiment, John uses a mesh to separate soil particles from water. Which technique of separation is he using?
Aneli [31]
Salutations!

<span>In a laboratory experiment, John uses a mesh to separate soil particles from water. Which technique of separation is he using?

The technique that John is using is the filtration technique. Filtration is a technique to separate the solid which is insoluble from the liquid. For instance: Sand and water, sand is insoluble, thus it stays in the filter paper, while the water proceeds through the filter paper.

Hope I helped :D</span>
3 0
3 years ago
The boiling point of a solution with a non-volatile solute is higher than that of the pure solvent . True or false
RUDIKE [14]
True because
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6 0
2 years ago
While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°
Lana71 [14]

Answer:

Final temperature  = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 -  25°C

305 cal = 5.25cal/°C × T2 -  25°C

305 cal / 5.25cal/°C = T2 -  25°C

58.1 °C = T2 -  25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0
2 years ago
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