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wel
3 years ago
10

Which of the following represents the molecules of one compound and atoms of one element?

Chemistry
1 answer:
ankoles [38]3 years ago
3 0

Answer:

i think its b.c but I'm not sure

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In the cases where no electric current flowed, do compounds form atoms, molecules or lattices in solution?
dsp73

Answer: 28

Explanation:a=d+a

8 0
2 years ago
So2+o2 = so3 método de tanteo​
gulaghasi [49]

Answer:

I dont understand

Explanation:

818191919

7 0
3 years ago
A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g ×\frac{1mol}{159,609g} = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g ×\frac{1mol}{18,02g} = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>

I hope it helps!

5 0
3 years ago
Which statement best describes visible light?
tekilochka [14]

Answer:

The Visible Spectrum Visible light is the light that we can see, and thus is the only light detectable by the human eye. White light is visible light, and it contains all the colors of the rainbow, from red to violet. The range of visible wavelengths is 400 to 700 nanometers.

Explanation:

6 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
3 years ago
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