What are the reactants for cellular respiration

Liquid A has a density of 0.75 g/ m and liquid B has a density of 1.14 g/mL. What will happen when both liquids are poured into

Ivan

Liquid A would rest on top of liquid B
Liquid B is much more dense and would sink to the bottom of the container wile liquid A would rest on top of it. Like how a pool noodle floats because it’s less dense than water other liquids can float as well.

5 0

3 years ago

Which of the intermolecular forces below occurs between polar molecules?

motikmotik

Answer:

dipole-dipole interactions

Explanation:

Dipole-dipole forces are the attractive forces that occur between polar molecules.

6 0

3 years ago

A 0.630 gram sample of a metal, M, reacts completely with sulfuric acid according to:A volume of 291 mL of hydrogen is collected

Lina20 [59]

Answer:

55.0 g/mol

Step-by-step explanation:

Step 1. Partial pressure of hydrogen

You are collecting the gas over water, so

$p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}$

$p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}$

$p_{\text{atm}} = \text{756.0 Torr}$

At 25 °C, $p_{\text{H}_{2}\text{O}} = \text{23.8 Torr}$

$p_{\text{H}_{2}} = \text{756.0 Torr} - \text{23.8 Torr} = \text{732.2 Torr}$

===============

Step 2. Moles of H₂

We can use the Ideal Gas Law.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

p = 732.2 Torr Convert to atmospheres

p = 732.2/760

p = 0.9634 atm

V = 291 mL Convert to litres

V = 0.291 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 25 °C Convert to kelvins

T = (25 + 273.15 ) K = 298.15 K

n = (0.9632 × 0.291)/(0.082 06 × 298.15)

n = 0.2804/24.47

n = 0.011 46 mol

===============

Step 3. Moles of metal

The partial chemical equation is

M + H₂SO₄ ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.011 46× 1/1

Moles of M = 0.011 46 mol M

===============

Step 4. Atomic mass of M

Atomic mass = mass of M/moles of M

Atomic mass = 0.630/0.011 46

Atomic mass = 55.0 g/mol

7 0

4 years ago

Calculate the percent ionization of a 0.18 M benzoic acid solution in a solution containing 0.10 M sodium benzoate.

Yuliya22 [10]

Answer:

% I = 0.083 %

Explanation:

C6H5COOH + NaOH ↔ NaC6H5CO2 + H2O

∴ M C6H5COOH = 0.18 mol/L

∴ M NaC6H5CO2 = 0.10 mo/L

% ion = ( { H3O+ ] / initial acid concentration ) * 100

⇒ C6H5COOH ↔ H3O+ + C6H5COO-

⇒ NaC6H5CO2 ↔ Na+ + C6H5COO-

∴ Ka = 6.4 E-5 = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ]

Ka value taken from the literature

mass balance

⇒ 0.18 + 0.1 = [ C6H5COOH ] + [ C6H5COO- ]

⇒ [ C6H5COOH ] = 0.28 - [ C6H5COO- ] ........(1)

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ C6H5COO- ]

∴ [ Na+ ] ≅ M NaC6H5CO2 = 0.1 M

⇒ [ C6H5COO- ] = [ H3O+ ] + 0.1..............(2)

(1) and (2) in Ka:

⇒ 6.4 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.1 ) / ( 0.28 - ( [ H3O+ ] + 0.1 ) )

⇒ 6.4 E-5 = [ H3O+ ]² + 0.1 [H3O+ ] / ( 0.18 - [ H3O+ ] )

⇒ 1.17 E-5 - 6.5 E-5 [ H3O+ ] = [ H3O+ ]² + 0.1 [ H3O+ ]

⇒ [ H3O+ ]² + 0.1 [ H3O+ ] - 1.17 E-5 = 0

⇒ [ H3O+ ] = 1.493 E-4 M

⇒ % I = ( 1.493 E-4 / 0.18 ) * 100 = 0.083 %

6 0

3 years ago

The rate of decay of a chemical involved in a reaction that is second order (bimolecular) in one reactant A is given by: = k [AJ

anyanavicka [17]

Explanation:

$2A\rightarrow products$

According to mass action,

$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2$

Where, k is the rate constant

So,

$\dfrac{d[A]}{dt}=-k[A]^2$

Integrating and applying limits,

$\int_{[A_t]}^{[A_0]}\frac{d[A]}{[A]^2}=-\int_{0}^{t}kdt$

we get:

$\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Half life is the time when the concentration reduced to half.

So, $[A_t]=\frac{1}{2}\times [A_0]$

Applying in the equation as:

$t_{1/2}=\dfrac{1}{k[A_o]}$

7 0

4 years ago