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3 years ago

In beta decay, what is emitted?

1 answer:

5 0

In beta minus (β−) decay, a neutron is converted to a proton and the process creates an electron and an electron antineutrino; while in beta plus (β+) decay, a proton is converted to a neutron and the process creates a positron and an electron neutrino. β+ decayis also known as positron emission.

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What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 = [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

7 0

3 years ago

( Cooking Popcorn using a mircowave Burning yourself by touching burning water the handle of a pot becoming to hot to touch as i

mihalych1998 [28]

Conduction - touching hot pot
Convection - oven cycling warm air
Conduction - touching a warm coffee mug
Heat from a fire - radiation
Heat from the sun to solar panel - radiation
Warm water rising - convection

3 0

3 years ago

A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the iden

Sindrei [870]

Answer:

HCl

Explanation:

Choices:

CO: 28.01g/mol

NO₂: 46g/mol

CH₄: 16.04g/mol

HCl: 36.4g/mol

CO₂: 44.01g/mol

It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:

PV = nRT

Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).

Replacing:

PV = nRT

PV / RT = n

0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n

8.786x10⁻³ = moles of the gas

As the mass of the gas is 0.320g; its molar mass is:

0.320g / 8.786x10⁻³moles = 36.4 g/mol

Based in the group of answer choices, the identity of the gas is:

4 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago

In an electric current, the ___ particles flow or are pushed along a path

MakcuM [25]

The negative particles flow or are pushed along a path

7 0

1 year ago