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3 years ago

Name one danger associated with blizzards

2 answers:

6 0

Blizzards can be dangerous on roads because the snow makes it difficult to see the road. Also blizzards can cause damage to houses.

horsena [70]3 years ago

5 0

Blizzards make it difficult to see on the road when you are driving.

You might be interested in

Write and balance the half-reaction for the oxidation of white phosphorous P4 to the phosphate ion PO3^−4 in a basic solution.

aalyn [17]

Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.

P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻

Final reaction :

P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.

The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).

Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.

For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).

Learn more about Half reactions here : brainly.com/question/2491738

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3 0

2 years ago

4. A student is doing experiments with CO2(q) . Originally a sample of the gas is in a rigid container at 299 Kand 0.70 atm. The

marta [7]

Answer:

b. 0,99atm

c. Answer is in the explanation

d. Answer is in the explanation

Explanation:

b. Using Gay-Lussac's law:

P₁T₂ = P₂T₁

P₁: 0,70 atm; T₂: 425K; P₂: ??; T₁: 299K

0,70atm×425K / 299K = 0,99 atm

c. Using kinetic molecular theory, the increasing of temperature increases the kinetic energy of gas particles and if kinetic energy increases, the pressure increases. That means the increasing of temperature increases the pressure in the system.

d. Now, the increases in kinetic energy of gases increase the collisions betwen particles. As these intermolecular forces that are not taken into account in ideal gas law, the observed pressure will be different to the pressure predicted by ideal gas law.

I hope it helps!

8 0

3 years ago

The following reaction plays a key role in the destruction of ozone In the atmosphere:

lys-0071 [83]

Answer:

19.91 J/K

Explanation:

The entropy is a measure of the randomness of the system, and it intends to increase in nature, thus for a spontaneous reaction ΔS > 0.

The entropy variation can be found by:

ΔS = ∑n*S° products - ∑n*S° reactants

Where n is the coefficient of the substance. The value of S° (standard molar entropy) can be found at a thermodynamic table.

S°, Cl(g) = 165.20 J/mol.K

S°, O3(g) = 238.93 J/mol.K

S°, O2(g) = 205.138 J/mol.K

So:

ΔS = (1*205.138 + 1*218.9) - (1*165.20 + 1*238.93)

ΔS = 19.91 J/K

6 0

3 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago

How much energy is required to change the temperature of 50g of water from 20 C to 55 C?

bagirrra123 [75]

You beef go have this much energy 60g

6 0

3 years ago