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Lilit [14]
3 years ago
15

What is the number of electrons and protons can be seen in this atom?

Chemistry
1 answer:
pochemuha3 years ago
8 0

Explanation:

the number of electrons = 8

and protons = 8 can be seen in this atom

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How is laughing gas made first answer gets brainleist and 100 points
Ket [755]

Answer:

Make nitrous oxide or laughing gas by heating ammonium nitrate and collecting the vapor by bubbling it up into a container over water. ... It's easy to make nitrous oxide or laughing gas at home or in the lab. All you need is a heat source and ammonium nitrate

Explanation:

5 0
2 years ago
Read 2 more answers
A beaker containing 6.32 moles of PBr3, contains___
NISA [10]

Answer:

3.8 x 10²⁴molecules

Explanation:

Given parameters:

Number of moles  = 6.32moles

Unknown:

Number of molecules  = ?

Solution:

The number of moles can be used to derive the number of molecules found within a substance.

Now,

       1 mole of substance contains 6.02 x 10²³ molecules

     6.32 mole of PBr₃ will contain 6.32 x 6.02 x 10²³ = 3.8 x 10²⁴molecules

8 0
3 years ago
What is the volume if 1.5 mol of gas has a pressure of 700 torr at 15°C?
Tasya [4]

Answer:

V = 38.48 L

Explanation:

Given that,

No. of moles = 1.5 mol

Pressure, P = 700 torr

Temperature, T = 15°C = 288 K

We need to find the volume of the gas. The ideal gas equation is given by :

PV=nRT\\\\V=\dfrac{nRT}{P}\\\\V=\dfrac{1.5\times 62.36\times 288}{700}\\\\V=38.48\ L, R = L.Torr.K⁻¹.mol⁻¹

So, the required volume is equal to 38.48 L.

7 0
3 years ago
If u answer this correctly I’ll mark you brainliest
olga2289 [7]

Answer:

6 is the right answer I know cause I like science

4 0
3 years ago
Read 2 more answers
A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f
Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0
3 years ago
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