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3 years ago

How would the world be different if people were not aware of radioactive decay

Chemistry

1 answer:

allsm [11]3 years ago

6 0

Answer:

everyone would die

Explanation:

if we did not know about it we would not do anything about it

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The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without jus

ioda

Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k = (2.303 /t) log [A₀] /[Aₙ]

where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>

[A₀] = 615 calories;

[Aₙ] = 615 - 480 = 135 calories

k = 2.00 * 10⁻²⁰ sec⁻¹

substituting the values in the equation of the rate constant;

2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)

(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)

t = 2.303 / 3.037 * 10⁻²⁰

t = 7.58 * 10¹⁹ seconds

8 0

3 years ago

HRISTINA HERRERA: Attempt 1

Lunna [17]

Answer:

7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.

Explanation:

Here we have the check if the mass of the reactants is equal to the mass of the products.

Reactants

$7.5+50=57.5\ \text{g}$

Products

$57.5\ \text{g}$

The data is consistent with the law of conservation of matter.

Reactants

$50+243=293\ \text{g}$

Products

$206+97=303\ \text{g}$

The data is not consistent with the law of conservation of matter.

Reactant

$17.7+34.7=52.4\ \text{g}$

Products

$62.4\ \text{g}$

The data is not consistent with the law of conservation of matter.

Only the first data is consistent with the law of conservation of matter.

8 0

3 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.</em>

<em />

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

<em>The moles of CH3COO- are its initial moles - the moles of HCl added</em>

<em>And moles of CH3COOH are its initial moles + moles HCl added</em>

<em />

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

<em />

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago

For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are

34kurt

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

8 0

3 years ago

How much mass of sodium chloride ( ) should be dissolved in water to make 1.5 L of 0.75 M aqueous solution? The molar mass of is

iogann1982 [59]

Answer:

Mass = 65.8 g

Explanation:

Given data:

Mass of sodium chloride = ?

Volume of solution = 1.5 L

Molarity of solution = 0.75 M

Solution:

Number of moles of sodium chloride:

Molarity = number of moles / volume in L

By putting values,

0.75 M = number of moles = 1.5 L

Number of moles = 0.75 M × 1.5 L

Number of moles = 1.125 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1.125 mol × 58.5 g/mol

Mass = 65.8 g

4 0

3 years ago