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2 years ago

How are the number of protons and neutrons represented in a periodic table?

Chemistry

1 answer:

sp2606 [1]2 years ago

5 0

Answer:

The atomic number or proton number (symbol Z) of a chemical element is the number of protons found in the nucleus of every atom of that element. The atomic number uniquely identifies a chemical element.

Explanation:

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Atoms of noble gases are generally inert because..

umka21 [38]

Answer:

B. their outer electron levels are filled

4 0

2 years ago

A helium balloon with a volume of 5.0 L and a pressure of 105 kPa at 25°C floats up high in the sky where the pressure is 850 mm

aalyn [17]

Answer:

volume 2- 4.011 L

8 0

2 years ago

Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kin

Dennis_Churaev [7]

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

$\mathsf{In \dfrac{C}{C_o} = -kt}$

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

$\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}$

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

3 0

3 years ago

A 232 mL cup of whole milk contains about 27 mg of cholesterol. Express the cholesterol concentration of the milk in kilograms p

son4ous [18]

Explanation:

Cholesterol concentration = 27 mg/ 232 mL

Change unit mg/ml to kg/m³

1 kg = 1000 g

1 g = 1000 mg

1 kg = 1000000 mg

1 kg = 10⁶ mg

10⁶ mg = 1 kg

1 mg = 10^-6 kg

m³ = 1000 L

1 L = 1000 mL

m³ = 1000000 mL

m³ = 10⁶ mL

10⁶ mL = m³

1 mL = 10^-6 m³

Cholesterol concentration = 27 mg/ 232 mL

27 × 10^-6 kg/ 232 × 10^-6 m³

= 27/232 kg/m³

6 0

3 years ago

The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial

Studentka2010 [4]

<u>Answer:</u> The partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

<u>Explanation:</u>

We are given:

$K_c=4.63\times 10^{-3}$

$p_{COCl_2}=0.760atm$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $4.63\times 10^{-3}$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $527^oC=527+273=800K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304$

The chemical reaction for the decomposition of phosgene follows the equation:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

At t = 0 0.760 0 0

At $t=t_{eq}$ 0.760-x x x

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

We are given:

$K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x$

Putting values in above equation, we get:

$0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656$

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

$p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm$

Hence, the partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

6 0

3 years ago