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yawa3891 [41]
2 years ago
15

Safety.

Chemistry
1 answer:
Ratling [72]2 years ago
8 0

Answer:

because gravitational force effects on wood as compare to paper

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Which type of energy would most likely create the worst environmental disaster from an accident or spill?
rewona [7]

Answer:

nuclear energy

Explanation:

6 0
2 years ago
30. a. How do the properties of metals differ from those
Ahat [919]

Explanation:

Metals are elements that ionized by loss of electrons.

Ionic and molecular compounds are usually non-metals.

Properties of metals:

  • Metals have free mobile electrons and the metallic bonding ensures that.
  • They are usually electropositive and freely looses their electrons.
  • None of the metal is soluble without a chemical change occurring.
  • They are ductile and malleable.
  • Metals are good conductor or heat and electricity in their free uncombined state.
  • They are lustrous.

B. The specific property of metals accountable for their unusual electrical conductivity is due to the presence of free mobile electrons in their lattices.

learn more:

Metals brainly.com/question/2474874

#learnwithBrainly

3 0
3 years ago
A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doin
melomori [17]

Answer:

ΔE = 73 J

Explanation:

By the first law of thermodynamics, the energy in the system must conserved:

ΔE = Q - W

Where ΔE is the internal energy, Q is the heat flow (positive if it's absorbed by the system, and negative if the system loses heat), and W is the work (positive if the system is expanding, and negative if the system is compressing).

So, Q = + 551 J, and W = + 478 J

ΔE = 551 - 478

ΔE = 73 J

3 0
3 years ago
To investigate the reactivity of metals a student uses four metals. Each time they added 1 g of the metal to 25 cm³ of sulfuric
Mazyrski [523]

Answer:

Volume of the sulfuric acid (25cm³), same mass of each metal (1g)

Explanation:

In an experiment, the CONTROL VARIABLE also known as constant is the variable that is kept unchanged for all groups in an experiment. This is done in order not to influence the outcome of the experiment.

In this case, students are trying to investigate the reactivity of four different metals. They added 1 g of each metal to 25cm³ of sulfuric acid and recorded the temperature change. Based on the explanation of control variable above, the VOLUME OF SULFURIC ACID (25cm³) and the MASS OF EACH METAL (1g) are the CONTROL VARIABLES because they are the same or unchanged in this experiment.

5 0
2 years ago
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