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DiKsa [7]
3 years ago
5

How many core electrons does argon have?

Chemistry
1 answer:
Artemon [7]3 years ago
4 0
10 core electrons are in argon.
You might be interested in
What branch of chemistry studies the flow of electrons?
grin007 [14]

Answer:

Your answer is B, Electrochemistry!

Explanation:

This is the part of chemistry that studies the chemical process in which electrons flow. This flow is called electricity. Electricity is generated by the flow of electrons, from one element to another element. This reaction is called oxidation reduction.

6 0
2 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
3 years ago
A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea
jekas [21]

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:

Q = m\cdot c \cdot \Delta T (1)

Where:

  • m - Mass of the sample, in grams.
  • c - Specific heat of water, in calories per gram-degree Celsius.
  • \Delta T - Temperature change, in degrees Celsius.

If we know that m = 10\,g, c = 1\,\frac{cal}{g\cdot ^{\circ}C} and \Delta T = 9\,^{\circ}C, then the heat released by the fuel is:

Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

7 0
2 years ago
Solve fasttt<br><br> C7H17+O2=CO2+H2O<br><br> Balancing Equations
sertanlavr [38]

Answer:

The answer to your question is

                                        4C₇H₁₇  + 45 O₂    ⇒   28 CO₂   +  34H₂O  

Explanation:

Write the equation

                                        C₇H₁₇  +    O₂    ⇒    CO₂   +    H₂O  

Process

1.- Check if the equation is balanced

                                 Reactants            Element              Products

                                        7                          C                           1

                                       17                          H                          2

                                        2                          O                          3

As the number of reactants and products is different, we conclude that the reaction is unbalanced.

2.- Write a coefficient "7" to CO₂   and a coefficient of 17/2 to H₂O

                                      C₇H₁₇  +    O₂    ⇒   7CO₂   +  \frac{17}{2}H₂O  

                                 Reactants            Element              Products

                                        7                          C                           7

                                       17                          H                          17

                                        2                          O                          51/2

3.- Write a coefficient of 45/2 to the O₂, and multiply all the equation by 2.

                         4C₇H₁₇  + 45 O₂    ⇒   28 CO₂   +  34H₂O  

                  Reactants            Element              Products

                        28                          C                        28

                        68                          H                        68

                        90                          O                        90

5 0
3 years ago
_______ are prophylactic agents used to treat bronchoconstriction.
Brut [27]
Bronchodilators <span>are prophylactic agents used to treat bronchoconstriction.</span>
6 0
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