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Svetach [21]
3 years ago
10

Please help. balance this equation

Chemistry
1 answer:
Nuetrik [128]3 years ago
4 0

Answer:

k=2

ci=2

o=6

k=2

ci=2

o=6

Explanation:

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Which of the following is TRUE? A) The equivalence point is where the amount of acid equals the amount of base during any acid-b
liberstina [14]

Answer:

the correct option is B

Explanation:

The correct option is b, since if we reach pH 7, it means that the acid-base reaction is neutralized, therefore the base has been neutralized by an acid or vice versa, without taking into account the proteins or the amounts of both components .

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3 years ago
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Predict whether each of the following is soluble in water or not.
wlad13 [49]

Answer:

The soluble in water are a and b. c is not soluble in water

Explanation:

The water solubility of a compound, it depends its polarity (. KHSO4 has several polar molecules, actually it has ions charged positively like K and H, and charged negatively like sulfate (SO4-2). When it happens, molecules can do hydrogen bridges with water. It is because of above, KHSO4 can be used in wine making. Something similar happens with propylen glycol but in this case the polar part is OH- ions. Benzene has not polar parts. It is totally apolar compound.

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4 0
4 years ago
An X-ray beam with λ = 157 pm incident on the surface of a crystal produced a maximum reflection at an angle of θ = 28.5 ∘. Assu
WARRIOR [948]

Answer:

The separation between layers of atoms in the crystal is 164.52 pm.

Explanation:

Formula used

n \lambda = 2d\sin \theta (Bragg's law)

Where :

n = an order of reflection (integer)

\lambda = wavelength of the beam incident X-ray

d = spacing between the layer

θ = incident angle ( angle between ray and scatter plane)

We have :

λ = 157 pm

θ = 28.5°

n = 1

d = ?

1\times 157 pm= 2d\sin (28.5^o) (Bragg's law)

d=\frac{1\times 157 pm}{2\times \sin (28.5^o)}=164.52 pm

The separation between layers of atoms in the crystal is 164.52 pm.

8 0
3 years ago
Determine the empirical formulas for compounds with the following percent compositions:
Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

3 0
3 years ago
Which is an unreliable source?
son4ous [18]
A source with opinions, no factual information.
8 0
3 years ago
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