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aleksandr82 [10.1K]
3 years ago
11

Explain and Describe the ingredients to create a storm and what factors determine how small or large it will be?

Chemistry
1 answer:
anygoal [31]3 years ago
8 0

Answer:

Explanation:

A lymphocyte is a type of white blood cell that is part of the immune system. There are two main types of lymphocytes: B cells and T cells. The B cells produce antibodies that are used to attack invading bacteria, viruses, and toxins.

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How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
OLEGan [10]

Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>

<em />

We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

3 0
3 years ago
What advice would you give to a person who handles hydrogen and oxygen in their workplace?I think I have an idea of what advice
finlep [7]
Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.
3 0
3 years ago
Calculate (a) the pH of a 0.0250 M solution of phenylacetic acid, and (b) the pH of a 0.0500 M solution of sodium phenylacetate.
almond37 [142]

Answer:

Explanation:cool thre answer is 009000000000.1 because you have to add them so ye

6 0
3 years ago
If the density of an object is 5 g/cm3 and it takes up 7 cm3 of space, calculate the mass?
statuscvo [17]

Answer:

<h2>The answer is option A</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of object = 7 cm³

density = 5 g/cm³

The mass of the object is

mass = 5 × 7

We have the final answer as

<h2>35 g</h2>

Hope this helps you

8 0
3 years ago
How many moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 ATM
Rama09 [41]

Answer:

285.4 moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 200.8 atm
  • V= 35 L
  • n=?
  • R= 0.082 \frac{atm*L}{mol* K}
  • T= 27.3 C= 300.3 K (being O C= 273 K)

Replacing:

200.8 atm* 35 L= n* 0.082\frac{atm*L}{mol* K} * 300.3 K

Solving:

n=\frac{200.8 atm* 35 L}{0.082\frac{atm*L}{mol* K} * 300.3 K}

n= 285.4 moles

<u><em>285.4 moles of gas are in a 35.0 L scuba canister if the temperature of the canister is 27.3 °C and the pressure is 200.8 atm.</em></u>

<u><em></em></u>

5 0
3 years ago
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