What does the symbol (–Delta.Hfus) indicate in a phase change?

Which type of radioactive decay occurs in the following reaction?

ella [17]

This one is beta decay (the -1 subscript tells us that)

7 0

3 years ago

Which statement is correct regarding the reaction below? 3A + 2B yields C + 2D The rate of formation of D is twice the rate of d

Vedmedyk [2.9K]

Answer:

The correct statements are:

The rate of disappearance of B is twice the rate of appearance of C.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

3A + 2B → C + 2D

Rate of the reaction:

$R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}$

$-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}$

$-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}$

The rate of disappearance of B is twice the rate of appearance of C.

$\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}$

$2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}$

8 0

3 years ago

Read 2 more answers

Which type of radiation involves the conversion of a neutron into a proton?

romanna [79]

Answer:

B. Beta

Explanation:

Neutron -> proton = responsible by Beta particles.

3 0

1 year ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

g The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures in excess of 1100 K. We can conclude that a. ΔG° is ne

mel-nik [20]

Answer:

c. ΔH° is positive and ΔS° is positive.

Explanation:

Hello,

In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

$\Delta G=\Delta H-T\Delta S$

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.

Best regards.

7 0

3 years ago