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4 years ago

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the diagram below represents two beakers, each containing an ice cube and clear liquid. in beaker A the ice cube floats, and in

beaker B the ice cube rest on the bottom. what is the most probable cause for the difference in behavior of the ice cube in the two beakers?

1 answer:

Mama L [17]4 years ago

4 0

Answer:

The density of the liquid in beaker B is less than the that of ice.

Explanation:

Ice will float if its mass is less than the mass of the liquid it displaces.

For example, the density of ice is less than that of water.

A 10 cm³ cube of ice has a mass of about 9 g, while the mass of 10 cm³ of water is 10 g. Thus, 9 g of ice displaces 10 g of water.

The denser water displaces the lighter ice and the ice floats to the top.

If the density of the liquid is <em>less than</em> that of water, say, 8 g/cm³, the ice will displace only 8 g of the liquid. The ice will sink.

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What will the presence of H+ ions in a solution cause?

vodka [1.7K]

The H+ will react with Magnesium to release hydrogen Gas under the equation Mg + 2H+ --> Mg(2+) + H2(g)

Hope this Helps! :)

8 0

3 years ago

Read 2 more answers

Which of the following is not a conjugate acid-base pair? A) NH4+/NH3 B) H30-OH OC) H2SO3/HSO3 D) C2H302-/HC2H302

raketka [301]

Answer : The option (d) is not a conjugate acid-base pair.

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

(a) The equilibrium reaction will be,

$NH_4^++OH^-\rightleftharpoons NH_3+H_2O$

In this reaction, $NH_4^+/NH_3$ are act as a conjugate acid-base.

(b) The equilibrium reaction will be,

$H_3O^++OH^-\rightleftharpoons H_2O+H_2O$

In this reaction, $H_3O^+/OH^-$ are act as a conjugate acid-base.

(c) The equilibrium reaction will be,

$H_2SO_3+OH^-\rightleftharpoons HSO_3^-+H_2O$

In this reaction, $H_2SO_3/HSO_3^-$ are act as a conjugate acid-base.

(d) The equilibrium reaction will be,

$C_2H_3O_2^-+H_2O\rightleftharpoons HC_2H_3O_2+OH^-$

In this reaction, $C_2H_3O_2^-/HC_2H_3O_2$ are act as a conjugate base-acid.

Hence, from this we conclude that, the option (d) is not a conjugate acid-base pair but it is a act as conjugate base-acid.

6 0

3 years ago

An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density

V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

$Density = \dfrac{4 \times M_w}{N_A \times a^3}$

where

$M_w$ = molar mass of the compound

$N_A=$ avogadro's constant

$a^3 =$ the volume of a unit cell

Given that:

Density $(\rho)$ = 19.30 g/cm³

a = 0.408 nm

a = $0.408 \times 10^{-9} \times 10^{2} \ cm$

a = $4.08 \times 10^ {-8} \ cm$

∴

$19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}$

$M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}$

$M_w=197.37 \ g/mol$

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago

If you need to produce X-ray radiation with a wavelength of 1 Å. a. Through what voltage difference must the electron be acceler

earnstyle [38]

Answer:

12.4×10^3 V

Explanation:

From E=hc/wavelength= eV

The voltage becomes

V= hc/e* wavelength

V= 6.63*10^-34*3*10^8/1.6*10^-19*1*10^-10

Note that the energy of the photon is transferred to the electron. That is the basic assumption we have applied in solving this problem. The kinetic energy of the electron is equal to the product of the electron charge and the acceleration potential.

4 0

4 years ago