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3 years ago

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When naming acids such as HCl, HBr, and HI, the prefix "hydro-" is added and the "ine" of the element is replaced with which of

the following suffixes before the word "acid" is added?
Select one:
a. "-ic"
b. "-ate"
c. "-ite"
d. "-ide"

1 answer:

liberstina [14]3 years ago

8 0

Answer:

a. "-ic"

Explanation:

The correct answer is a. "-ic".

For example:

For HCl, the name is Hydrochloric acid. As you can see, the suffix "-ine" of the element (originally chlor<em>ine</em>) is changed to "-ic".

Following the same logic, the name for HBr is Hydrobromic acid; and the name for HI is Hydroiodic acid.

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Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

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Answer:

The answer to your question is Only A and B

Explanation:

Reactions

A. 3 AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃

Reactants Elements Products

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3 Br 3

1 Ga 1

1 P 1

4 O 4

B. 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O

Reactants Elements Products

3 S 3

2 B 2

12 H 12

18 O 18

C. Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag

Reactants Elements Products

1 Fe 1

2 Ag 2

2 N 2

6 O 6

D. C₂H₄O₂ + 2O₂ → 2CO₂ + 2H₂O

Reactants Elements Products

2 C 2

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4 years ago

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Deploying an airbag in an automobile is a complex process. Identify at least three steps in the process. Refer to your research

gavmur [86]

Answer:

1) The car must have an accelerometer and airbag circut.

2) Chemical explosive of enough sodium azide to fill the bag when ignited

3) Airbag can be covered in talcum powder to help it unwrap smoothly.

Explanation:

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You have a 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

lions [1.4K]

Answer:

$\boxed {\boxed {\sf 0.4 \ mol \ HCl}}$

Explanation:

Molarity is concentration measured in moles per liters. It is the number of moles of solute per liters of solution. The formula is:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}}$

We know the solution of HCl has a molarity of 0.5 molar and there are 0.8 liters of solution.

1 molar (M) is equal to 1 mole per liter.
Let's convert the molarity of 0.5 M HCl to 0.5 mol HCl per liter. This will make unit cancellation easier.

The moles of solute or HCl are unknown, so we can use x. Now, we can substitute all known values into the formula.

$0.5 \ mol \ HCl /L = \frac {x}{0.8 \ L}$

Since we are solving for the moles of solute (x), we must isolate the variable. It is being divided by 0.8 liters. The inverse of division is multiplication, so we multiply both sides by 0.8 L.

$0.8 \ L *0.5 \ mol \ HCl /L = \frac {x}{0.8 \ L} *0.8 \ L$

$0.8 \ L *0.5 \ mol \ HCl /L=x$

The units of liters (L) cancel.

$0.8 * 0.5 \ mol \ HCl= x$

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This solution contains <u>0.4 moles of HCl.</u>

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