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hammer [34]
3 years ago
5

Write the ions present in the solution of cuso4. express your answers as chemical formulas separated by a comma. identify all of

the phases in your answers.
Chemistry
2 answers:
Contact [7]3 years ago
8 0
Copper Sulfate (CuSO4) compound is basically comprised of Copper (II) ion and Sulfate ion. Ions are C<span>u2</span>+ and S<span>O4 2</span><span>−</span>.

<em>ANSWER:</em><em>Cu 2+ and SO4 2-</em>
Soloha48 [4]3 years ago
8 0

Answer: The ions which are present in the solution of CuSO_4 are Cu^{2+}\text{ and }SO_4^{2-}, both in aqueous state.

Explanation: When CuSO_4 is dissolved in water, it forms an aqueous solution. The solution contains two ions, both in aqueous states.

Equation follows:

CuSO_4(aq.)\rightarrow Cu^{2+}(aq.)+SO_4^{2-}(aq.)

Ions that are present in the solution of CuSO_4 are Cu^{2+}\text{ and }SO_4^{2-}

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Magnesium(?)
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Do non metals have lower melting points than metals?
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I WILL MARK BRAINLISET FOR WHOEVER GETS THIS RIGHT! A pot is placed on a gas flame, and the water inside the pot begins to boil.
NNADVOKAT [17]

Answer:

B.     Warm water rises within the pot.?

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8 0
3 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
2 years ago
(10 points ) please help this is major grade .
erastova [34]

Answer:

absorbed

Explanation:

because black absorbs the light around it

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2 years ago
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