All categories

3 years ago

The four fundamental forces are arranged in the increasing order of their relative strength.

Chemistry

1 answer:

natali 33 [55]3 years ago

3 0

Answer:

The correct answer is - both act only between non-atomic particles.

Explanation:

The decreasing order of their relative strength is - the strong force, electromagnetism, weak force, gravity.

so, A would be here = gravity

B would be = weak force

C would be = strong force

The weak and strong forces both are fundamental forces that act on sub-atomic particles only such as quarks.

You might be interested in

The combustion of hydrogen and oxygen is commonly used to

posledela

Answer:

6.66 mol

Explanation:

(atm x L) ÷ (0.0821 x K)

(0.875 x 250) ÷ (0.0821 x 400)

=6.66108

5 0

3 years ago

g Suppose a sample of an ideal gas in a container is subjected to a temperature change. A decrease in temperature will the kinet

Umnica [9.8K]

Answer:

323.15 °C

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.

P ∝ T

Also,

Using Charle's law

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

Given ,

P₂ = 2P₁

T₁ = 25 °C

T₂ = ?

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

Using above equation as:

$\frac{P_1}{298.15}=\frac{2P_1}{T_2}$

$T_2=2\times 298.15\ K$

New temperature = 596.3 K

Also,

T(K) - 273.15 = T( °C)

So, Temperature = 596.3 - 273.15 °C = 323.15 °C

5 0

3 years ago

Can anyone help me in doing this question?

Alecsey [184]

1. No. 108 g of Ag has a lower number of atoms than 80 g of Ca.

2. The gram of $SO_3$ in 3 moles of the compound would be 240 g.

3. 1.20 x $10^{24$ atoms

4. The grams of hydrogen in 3.6 x $10^{24$ molecules of $NH_3$ would be 18 grams.

<h3>Number of atoms in compounds</h3>

According to Avogadro, 1 mole of a substance contains 6.022 x $10^{23$ molecules or atoms.

Recall that: mole = mass/molar mass

1. 108 g of Ag would be equivalent to: 108/108 = 1 mol.

80 g of Ca would be equivalent to: 80/40 = 2 mol

Since 1 mol is equivalent to 6.022 x $10^{23$ molecules or atoms, it means 80 g of Ca has twice as atoms as 108 g of Ag.

2. 3 mol sample of $SO_3$ would be equivalent to: 3 x 80 = 240 g

3. 124 g of $Na_2O$ would be equivalent to: 124/62 = 2 mol

Number of atoms = 2 x 6.022 x $10^{23$

= 1.20 x $10^{24$ atoms

4. 3.6 x $10^{24$ molecules of $NH_3$ would be equivalent to:

3.6 x $10^{24$ /6.022 x $10^{23$ = 6 mol of $NH_3$

$NH_3 --- > 3H^+ + N^{3-$

From the above equation, 1 mole of $NH_3$ produces 3 moles of hydrogen. Thus, 6 moles of $NH_3$ would be equivalent to 18 moles of hydrogen.

18 moles of hydrogen = 18 x 1

= 18 g

More on the number of atoms in samples can be found here: brainly.com/question/28834341

#SPJ1

5 0

1 year ago

Hydrogen has three isotopes 1H, 2H, and 3H. What is the difference between these three isotopes? The number of protons ranges fr

adell [148]

The number of neutrons ranges from 0 to 2. By definition isotopes have the same number of Protons, but different number of Neutrons.

3 0

3 years ago

Read 2 more answers

The minerals hematite (Fe2O3) and magnetite (Fe3O4) exist in equilibrium with atmospheric oxygen:4Fe3O4(s) + O2(g) ⇌ 6Fe2O3(s)Kp

erastovalidia [21]

Answer:

For a: The partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

For b: The reaction must proceed in the forward direction to reach equilibrium.

For c: The value of $K_c$ is $6.12\times 10^{88}$

Explanation:

For the given chemical equation:

$4Fe_3O_4(s)+O_2(g)\rightleftharpoons 6Fe_2O_3(s)$

For a:

In the expression of $K_p$ , the partial pressures of solids and liquids are taken as 1.

The expression of $K_p$ for above equation follows:

$K_p=\frac{1}{p_{O_2}}$

We are given:

$K_p=2.5\times 10^{87}$

Putting values in above equation, we get:

$2.5\times 10^{87}=\frac{1}{p_{O_2}}\\\\p_{O_2}=\frac{1}{2.5\times 10^{87}}=4.0\times 10^{-88}atm$

Hence, the partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

For b:

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{1}{p_{O_2}}$

We are given:

$p_{O_2}=0.21$

Putting values in above equation, we get:

$Q_p=\frac{1}{0.21}=4.76$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction must proceed in the forward direction to reach equilibrium.

For c:

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 = $2.5\times 10^{87}$

$K_c$ = equilibrium constant in terms of concentration = ?

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

T = temperature = 298 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=0-1=-1$

Putting values in above equation, we get:

$2.5\times 10^{87}=K_c\times (0.0821\times 298)^{-1}\\\\K_c=6.12\times 10^{88}$

Hence, the value of $K_c$ is $6.12\times 10^{88}$

7 0

3 years ago