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Scorpion4ik [409]
2 years ago
7

Yo yuh guys know this

Chemistry
1 answer:
erastovalidia [21]2 years ago
8 0

Answer:80KM is distance. 30KM north is displacement.

Explanation:

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kiruha [24]

Answer:

y = 5cos(\frac{2\pi }{6} x)

Explanation:

You can put that equation into a program like desmos on your browser and take a screenshot or use windows snippet tool.

Equation of a Wave: y = Acos(((2*pi)/B)x)

or

y = Acos(\frac{2\pi }{B} x)

A = Amplitude

B= Wavelength

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Are frosted flakes healthier than fruit loops?
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Answer:

The answer is below!!

Explanation:

Although Frosted Flakes are higher in sugar than other cereals and often not considered diet food, they can be part of a healthy weight-loss plan. Cereals that are low in calories and fat content such as Frosted Flakes may help you lose weight.

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4 0
2 years ago
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When combining with nonmetallic atoms, metallic atoms generally will?
Svetradugi [14.3K]
Ionic bonding I think?
6 0
3 years ago
2. A sample of iron has the dimensions of 2.0 cm x 3.0 cm x
zepelin [54]
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8 0
3 years ago
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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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