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V125BC [204]
3 years ago
7

Symptoms of a mental disorder are ways of

Chemistry
1 answer:
kkurt [141]3 years ago
6 0

Answer: thinking

Explanation:

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9, What is an Independent
stepan [7]

Answer:

Not influenced or controlled by others in matters of opinion

8 0
3 years ago
How many moles of glucose (C6H12O6) are in 5.0 liters of a 2.5 M C6H12O6 solution?
zhannawk [14.2K]
N = M x V      

n = 2.5 x 5.0
 
n = 12.5 moles of C6H12O6
5 0
3 years ago
Need help with this can some please help
vekshin1
I believe you would just put a 2 in front of NH3 and keep the other ones as 1
7 0
3 years ago
hurry please! avogadro's law relates the volume of a gas to the number of moles of gas when temperature and pressure are constan
DIA [1.3K]

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:

V1/n1 = V2/n2

V/2 = 2V/n2

Cross multiply

V x n2 = 2 x 2V

Divide both side by V

n2 = (2 x 2V)/V

n2 = 2 x 2

n2 = 4 moles

Therefore, 4 moles of the gaseous product were produced.

5 0
3 years ago
The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what
Lena [83]

Answer:

B) 7.7

Explanation:

For the reaction    Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)

Kc = (CO₃²⁻) / (CrO₄²⁻)

and the Ksp given are

Ag₂CO₃    ⇒  2 Ag⁺(aq) + CO₃²⁻(aq)    Ksp₁ = (Ag⁺)²(CO₃²⁻)  

Ag₂CrO₄   ⇒  2 Ag⁺(aq)+ CrO₄²⁻(aq)   Ksp₂ = (Ag⁺)²(CrO₄²⁻)

Where (...) indicate concentrations M

Notice if we divide the expressions for Ksp we get:

Ksp₁/Ksp₂ = (CO₃²⁻)  / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7

which is the desired answer.

7 0
3 years ago
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