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3 years ago

Beth takes a sip of very hot soup and decides to put an ice cube in her bowl. Which best describes what happens next?

1 answer:

4 0

Conduction because the hot soup heats up the bowl causing the heat to be transferred from the soup to the bowl to your hand when you touch it.

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Which of the following Gila Monster adaptations is based on the reproductive cycles of Gambel’s quails and cottontail rabbits?

Natasha_Volkova [10]

Answer:

Explanation:

3 0

3 years ago

Suppose 110.0 mL110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen

Fittoniya [83]

Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now write the balanced chemical equation

H₂ + F₂ → 2HF

<h3>What is Ideal Gas ?</h3>

An ideal gas is a gas that obey gas laws at all temperature and pressure conditions. It have velocity and mass but do not have volume. Ideal gas is also called perfect gas. Ideal gas is a hypothetical gas.

It is expressed as:

PV = nRT

where,

P = Pressure

V = Volume

n = number of moles

R = Ideal gas constant

T = temperature

Here,

P = 1 atm [At STP]

V = 110 ml = 0.11 L

T = 273 K [At STP]

R = 0.0821 [Ideal gas constant]

Now put the values in above expression

PV = nRT

1 atm × 0.11 L = n × 0.0821 L.atm/ K. mol × 273 K

$n = \frac{1\ \text{atm} \times 0.11\ L}{0.0821\ \text{L. atm/ K. mol} \times 273\ K}$

n = 0.0049 mol

<h3>How to find the concentration of resulting solution ? </h3>

To calculate the concentration of resulting solution use the expression

$C = \frac{n}{V}$

$= \frac{0.0049}{0.15}$

= 0.032 M

Thus from the above conclusion we can say that Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.

Learn more about the Ideal Gas here: brainly.com/question/25290815
#SPJ4

4 0

2 years ago

A tank containing both hf and hbr gases developed a leak. The ratio of the rate of effusion of hf to the rate of effusion of hbr

Pavlova-9 [17]

Answer:

2.01

Explanation:

The effusion is the passage of the molecules by a small hole by a difference of pressure. By Graham's Law, the rate of the effusion is inversely proportional to the square of the molar mass of the compound. Thus,

rateHF/rateHBr = √MHBr /√MHF

MHBr = 81 g/mol

MHF = 20 g/mol

rateHF/rateHBr = √81/√20

rateHF/rateHBr = 2.01

4 0

3 years ago

What happens to an atom when it gains an electron?

jasenka [17]

I'm pretty sure it becomes an ion! If an atom gains a negative electron, it becomes an ion.

6 0

3 years ago

Read 2 more answers

Hydrogen sulfate reacts with water to produce sulfate ions and hydronium ions: hso4−+h2o⇌so42−+h3o+ identify the conjugate acid-

Angelina_Jolie [31]

The generic equation for a reaction between an acid and water is

$HX_{(aq)} + H_{2}O_{(aq)} --> X_{(aq)}^{-} + H_{3}O_{(aq)}^+$

When an acid "reacts" with water, water acts as the base that accepts the proton (H+) from the acid. The remaining ion that is formed after the acid has donated its proton is called the conjugate base ( $X^-$ ), and the conjugate acid-base pair is $HX$ - $X^-$ .

Hydrogen sulfate ( $HSO_{4}^-$ ) is an ion from sulfuric acid. It is still an acid in itself and can "react" with water (( $H_{2}O$ ) to form the sulfate ( $SO_{4}^2-$ ) and hydronium ( $H_{3}O^+$ )ions.

$HSO_{4(aq)}^{-} + H_{2}O_{(aq)} --> SO_{4(aq)}^{2-} + H_{3}O_{(aq)}^+$

Based on the previous discussion, $SO_{4(aq)}^2-$ is identified to be the conjugate of the acid $HSO_{4(aq)}^-$ . Thus, the conjugate acid-base pair is $HSO_{4(aq)}^{-} - SO_{4(aq)}^{2-}$ .

8 0

3 years ago