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3 years ago

What type of bonding is found in the molecule HBr?

1 answer:

4 0

D)polar covalent because ionic and metallics include metals and nonpolar covalent wouldn’t work because it’s lewis for structure and en doesn’t match its requirements.

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What happens in a reduction?

MatroZZZ [7]

Answer:

A. An element gains one or more electrons

Explanation:

Reduction is a chemical reaction that involves the gaining of electrons by one of the atoms involved in the reaction between two chemicals. The term refers to the element that

6 0

4 years ago

Read 2 more answers

A certain element forms an ion with 36 electrons and a charge of +2. identify the element. express your answer as a chemical sym

blsea [12.9K]

The element is Sr (strontium)

strontium is in atomic number 38 in the periodic table. Strontium has 2.8.8.8.8.2

it loses two electrons to become stable hence it has a charge of 2+. when strontium loses two electron it form ion with 36 electrons

6 0

3 years ago

A ____________ star would have the hottest surface temperature.

Alexandra [31]

Answer:

Blue

Explanation:

If you look at a flame, blue is always at the bottom right? So that would be common sense that blue would be the hottest.

7 0

3 years ago

Read 2 more answers

2. Consider the reaction 2 Cg H18 (4) +250â (9) âº 16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1

Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of $C_8H_{18}$ = 16.15 moles

First we have to calculate the moles of $H_2O$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react to give 18 moles of $H_2O$

So, 16.15 moles of $C_8H_{18}$ react to give $\frac{16.15}{2}\times 18=145.35$ moles of $H_2O$

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of $C_8H_{18}$ = 878 g

Molar mass of $C_8H_{18}$ = 114 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_8H_{18}$ .

$\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_8H_{18}$ react with 25 moles of $O_2$

So, 7.702 moles of $C_8H_{18}$ react with $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

Now we have to calculate the mass of $O_2$ .

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g$

The mass of oxygen needed are 3080.8 grams.

6 0

3 years ago

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:

Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

7 0

3 years ago