Ask question

Ask question

All categories

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

You might be interested in

Defferent between crysttiline solid and amerphus solid

Alexxx [7]

Answer:

Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Explanation:

Hope this helped!

5 0

3 years ago

Read 2 more answers

In the copper – silver nitrate lab Copper medals and silver nitrate solution reacted to produce silver metal and copper (II) nit

Nataly [62]

Answer:

Percent Yield = 97.75 %

Explanation:

1 MOLE = It is equal to the molar mass of the substance

1 mole of Cu = 63.54 g (Molar Mass of Cu = 63.54 g/mole)

1 mole of AgNO3 = 170 g (Molar Mass of AgNO3 = 170 g/mol)

Given Mass of AgNO3 = 1.41 g

Given Mass of Cu = 2.93 g

<em><u>Second step : Find the limiting Reagent (which is in less amount) </u></em>

Balanced Chemical equation :

$Cu + 2AgNO_{3} \rightarrow Cu(NO_{3})_{2}+2Ag$

This means

1 mole of Cu will react with = 2 mole of AgNO3

63.54 g of Cu reacts with = 2 x 170 g of AgNO3

1 g of Cu reacts with = (2 x 170)/63.54 of AgNO3

$= \frac{170\times 2}{63.54}$

= 5.35 g of AgNO3

2.93 g should reacts with = 2.93 x 5.35 = 15.67 g of AgNO3

Available AgNO3 = 1.41 g

So , AgNO3 is less than required = limiting reagent

Now the reaction occur 1.41 g of AgNO3

Now, Limiting reagent will decide How much Silver(Ag) Metal will form

2 mole of AgNO3 will produce = 2 mole of Ag

1 mole of AgNO3 will produce = 1 mole of Ag

170 g AgNO3 will produce = 107.86 mole of Ag(Molar mass of Ag = 107.86)

1 g AgNO3 will produce =

$\frac{107.86}{170}$

1.41 g of AgNO3 will produce =

$\frac{107.86\times 1.41}{170}$

= 0.89 g

$Yield\ Percent = \frac{Actual\ Yield}{Theoritical\ Yield}\times 100$

Actual yield = 0.87 g

Theoritical yield = 0.89 g

$Yield\ Percent = \frac{0.87}{0.89}\times 100$

Percent Yield = 97.75 %

5 0

3 years ago

How do you compare the nuclear membrane and the cell membrane?

iragen [17]

A. Both the cell membrane and the nuclear membrane are protective coverings.

5 0

3 years ago

Read 2 more answers

Nitrogen and hydrogen react to produce ammonia. What mass of ammonia could be produced from 500 grams of nitrogen? Assume that e

OlgaM077 [116]

The answer is 607g. the working is shown above.

5 0

3 years ago

What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?

Anni [7]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

Let's find the number of periods elapsed (x), let us see:

$m = \dfrac{m_o}{2^x}$

$5 = \dfrac{20}{2^x}$

$2^x = \dfrac{20}{5}$

$2^x = 4$

$2^x = 2^2$

$\boxed{x = 2}$

Now, let's find the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$8 = 2*P$

$2\:P = 8$

$P = \dfrac{8}{2}$

$\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

5 0

3 years ago