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Vladimir79 [104]
3 years ago
10

90 grams of NaNO3 were added to solution at 10 degrees Celsius. What type of solution was formed?

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

Answer:

friend me on here and imma send you the link Explanation:

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Between which two atoms of water are hydrogen bonds are formed?
JulsSmile [24]

Answer:

the hydrogen atom of one water molecule and the lone pair of electrons on an oxygen atom of a neighboring water molecule.

5 0
2 years ago
Name the following alkane molecule:
Zina [86]

Answer:

B

Explanation:

its upside down but I'm 99% sure

5 0
3 years ago
How many moles are there of a 7 M solution with a volume of 14.44 liters?
9966 [12]

moles = molarity * 1000/volume

7*1000/14.44= 484.76

6 0
3 years ago
Ag(No3)+Na3(Po4)+Ag3(Po4)+Na(No3)
musickatia [10]

3AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3NaNO₃

Explanation:

                         AgNO₃+Na₃(PO₄) → Ag₃(PO₄) + NaNO₃

To balance this chemical equation, we can adopt a simple mathematical approach through which we can establish simple and solvable algebraic equations.

           aAgNO₃ + bNa₃PO₄ → cAg₃PO₄ + dNaNO₃

a, b, c and d are the coefficients needed to balance the equation.

Conserving Ag:    a = 3c

                     N:      a = d

                     O:       2a + 4b = 4c + 2d

                     Na:      3b = d

                      P:        b = c  

let a = 1; d = 1

    b = \frac{1}{3}

    c = \frac{1}{3}

                         

Multiplying through by 3:

  a = 3, b = 1, c = 1 and d = 3

                3AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3NaNO₃

Learn more:

Balanced equation   brainly.com/question/5964324

#learnwithBrainly

5 0
3 years ago
If a solution of hf (ka = 6.8 10-4) has a ph of 3.67, calculate the total concentration of hydrofluoric acid.
m_a_m_a [10]
When PH = ㏒(H^+)
[H+] = 10^-3.67
[H+] = 2.14x 10^-4

so according to the reaction equation:
                               HF               ↔        H^+          +          F^-
at equilibrium   X-(2.14x10^-4)           (2.14x10^-4)         (2.14x10^-4)

by substitution in Ka formula:
       Ka      = [H+][F]-/[HF]
6.8x10^-4 = (2.14x10^-4)*(2.14x10^-4) /(X-2.4x10^-4)
X-2.4x10^-4 = (4.58x10^-8)* (6.8x10^-4)
∴X = 2.4x10^-4
∴[HF] = X- (2.14x10^-4)= (2.4x10^-4)-(2.14x10^-4) = 2.6 X 10^-5

3 0
3 years ago
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