Can u pls help me with this question

Which type of bond will most likely be found in HBr?

NARA [144]

HBr is covalent bond.!

Hydrogen and Bromine will share their electrons, to obtain the electron configuration of a noble gas.

hope this helps!

6 0

3 years ago

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Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

4 years ago

1. 2.0 mL/s x 4.0 hrs = ? L How do you do this conversion problem

Phantasy [73]

Answer:

29 L.

Explanation:

Hello!

In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:

$2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L$

Which turns out 29 L with 2 significant figures.

Best regards!

6 0

3 years ago

The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole

Oksanka [162]

Answer: The partial pressure of $Cl_2$ is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

The given balanced equilibrium reaction is,

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Pressure at eqm. 0.973 atm 0.548atm x atm

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}$

Now put all the given values in this expression, we get :

$1.05=\frac{(0.548)\times (x)}{(0.973)}$

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of $Cl_2$ is 1.86 atm

4 0

4 years ago

A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin

gtnhenbr [62]

Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)

The ksp is defined as:

ksp = [borate²⁻] [Na⁺]²

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl× $\frac{1molB_{4}O_{7}^{2-}}{2molHCl}$ = 3,00x10⁻³ mol of B₄O₇²⁻

For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺

The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M

And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

ksp = 0,176

I hope it helps!

8 0

3 years ago

Can u pls help me with this question ​

Can u pls help me with this question