What are cilia? smooth covering hair like structures a tail all of the above

One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the

Sidana [21]

Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

Change in heat (ΔH) = 150 joules

Temperature (T) = 150 K

Find:

ΔS surrounding (entropy change of the reservoir)

Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

3 0

4 years ago

Consider a simple reaction in which a reactant AA forms products: A→productsA→products What is the rate law if the reaction is z

levacccp [35]

Answer :

The rate law expression for zero order reaction will be:

$Rate=k[A]^0$

The rate law expression for first order reaction will be:

$Rate=k[A]^1$

The rate law expression for second order reaction will be:

$Rate=k[A]^2$

Zero order reaction : There is no affect on the rate law.

First order reaction : The rate law becomes doubled.

Second order reaction : The rate law becomes quadrupled.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given reaction is:

$A\rightarrow Products$

The rate law expression for zero order reaction will be:

$Rate=k[A]^0$

The rate law expression for first order reaction will be:

$Rate=k[A]^1$

The rate law expression for second order reaction will be:

$Rate=k[A]^2$

Now we have to determine that if doubling of the concentration of A then the rate of reaction will be:

As we know that the zero order reaction does not depend on the concentration of reactant. So, there is no affect on the rate law.

As we know that the first order reaction depend on the concentration of reactant. So, the rate law becomes doubled.

As we know that the second order reaction depend on the concentration of reactant. So, the rate law becomes quadrupled.

5 0

4 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (

Taya2010 [7]

Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

NaOH → Na+ + OH-

⇒ C NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ C NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ C HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ C NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ C HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ C HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ C HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

8 0

4 years ago

Read 2 more answers

A 0.435 g sample of a metal, M, reacts completely with sulfuric acid according to M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4 ( aq ) + H 2

Tresset [83]

Answer: The molar mass of metal is 52.4 g/mol

Explanation:

To calculate the moles of hydrogen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the hydrogen gas = Total pressure - vapor pressure of water = (756.0 - 23.8 ) torr = 732.2 torr

V = Volume of the gas = 201 mL = 0.201 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$732.2torr\times 0.210L=n\times 62.364\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\n=\frac{732.2\times 0.210}{62.364\times 298}=0.0083mol$

The given chemical equation follows:

$M(s)+H_ 2SO_4(aq.)\rightarrow MSO_4(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is formed by 1 mole of metal

So, 0.0083 moles of hydrogen gas will be formed by = $\frac{1}{1}\times 0.0083=0.0083mol$ of metal

To calculate the molar mass of metal from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of metal = 0.0083 moles

Given mass of metal = 0.435 g

Putting values in above equation, we get:

$0.0083mol=\frac{0.435g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{0.435g}{0.0083mol}=52.4g/mol$

Hence, the molar mass of metal is 52.4 g/mol

7 0

4 years ago

1. A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C. If the

Maurinko [17]

Answer:

Mass of aluminium in sample = 3.591 g ≅ 3.6 grams

Explanation:

Given that, A sample of aluminum absorbs 50.1 J of heat, upon which the temperature of the sample increases from 20.0°C to 35.5°C.

the specific heat of aluminum is 0.900 J/g- °C

The relation between heat absorbed and change in temperature is given by, Q = msΔT.

where Q = heat absorbed

m = mass of the substance

s = specific heat of substance

ΔT = change in temperature

Now, in our case, Q = 50.1 J ; s = 0.900 J/g- °C; ΔT= 35.5-20 = 15.5°C

⇒ m = $\frac{Q}{s(T_{2} -T_{1}) }$

⇒ m = $\frac{50.1}{0.900(15.5)}$ = 3.591 g ≅ 3.6 g

⇒ m ≅ 3.6 g

5 0

3 years ago