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1 year ago

11

What type of bonds are shown in this diagram?

2 answers:

Dmitrij [34]1 year ago

6 0

Explanation:

no diagram is available in this question , so I am sorry not for replying this answer

qaws [65]1 year ago

3 0

Answer:

What type of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) metallic bonds

In what type of bonds do atoms join together because their opposite charges attract each other?

A: metallic bonds and covalent bonds

B: metallic bonds and ionic bonds

C: ionic bonds and covalent bonds

D: ionic bonds and hydrogen bonds

(answer) ionic bonds and hydrogen bonds

What types of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) hydrogen bonds

Which statement best describes the types of bonds shown in the diagram?

A: an ionic bond; the hydrogen chloride molecule has an electrical charge

B: an ionic bond; a hydrogen ion is bonding with a chlorine atom

C: a covalent bond; the hydrogen atom’s two electrons are being shared with the chlorine atom

D: a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

(answer) a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

Which of the following bonds is the strongest?

A: hydrogen bonds

B: metallic bonds

C: valence bonds

D: covalent bonds

(answer)

Explanation:

UwU

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Determine the potential deaths resulting from the following exposure to chlorine:

jekas [21]

Answer:

For each scenario as following:

A. 3 Potential deaths by chlorine exposure

B. 1 Potential deaths by chlorine exposure

C. 3 Potential deaths by chlorine exposure

Explanation:

According to Freitag, 1941 Chlorine exposure can be lethal at the concentration of 34-51 ppm in a time of 1h-1.5h. The answers are based on his reference.

6 0

3 years ago

How many grams of silver nitrate are needed to react with 156.2g of sodium sulfide to produce 595.8g of silver sulfide and 340.0

qaws [65]

Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃

Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate

Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams

3 0

3 years ago

Will mark BRAINLIEST!! <br> PLEASE ANSWER I HAVE 5 MINS

maksim [4K]

Answer:

Explanation:

7A 0.2

7B cork

7C Yes because the lighter it is the more likely it will float

7D Density is one of the fundamental scientific principles of life. It can describe any everyday object. Despite its significance, students often struggle to understand what it really is. Density is a measurement of how much space or volume is packed in an object or substance.

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3 0

2 years ago

What is the formula unit for Li and O?

Margarita [4]

The formula unit for Li and O is —————-> Li2O

4 0

3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

3 years ago