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3 years ago

If you had a beaker full of black powder which of the following could you have?

2 answers:

8 0

A beaker of black powder can be a mixture with different substances (i.e gunpowder: a mixture of sulfur, charcoal, etc.), an element which can probably be carbon (known as black), and a compound as well which can be potassium nitrate. Hence, the answer is D.

kolbaska11 [484]3 years ago

4 0

Answer:

a. mixture

Explanation:

A black powder probably represents a mixture of several dark colored elements, or several elements that reacted and formed a black colored mixture.

Mixture is the union of two or more different substances (regardless of whether they are simple or compound). It has different and variable (non-fixed) physical characteristics (melting point, boiling point, density, toughness, etc.) compared to its constituent substances.

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what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c

slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

$Q = m \times C \times \bigtriangleup T$

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

$C = \frac{Q}{m \times \bigtriangleup T}$

Let us plug in the values in above equation.

$C = \frac{300J}{267g \times 12 C}$

$C = \frac{300J}{3204 g C}$

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0

3 years ago

Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:

Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and $1s^22s^22p^6$

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

This element will easily loose 1 electron and form $Rb^+$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

$1s^22s^22p^3$

This element will easily gain 3 electrons and form $N^{3-}$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

$1s^22s^22p^6$

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

This element will easily gain 1 electron and form $Br^-$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

4 0

3 years ago

What is the relationship between events and patterns?

MAXImum [283]

Events that happen over and over create/become a pattern

8 0

3 years ago

Read 2 more answers

Calculate the atomic mass for copper using the weighted average mass method. Express your answer using two decimal places and in

tatuchka [14]

Answer:

63. 55 amu

Explanation:

Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.

Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.

Similarly,

Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%

Therefore, using the weighted average mass method, the atomic mass of copper is:

Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu

Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu

8 0

3 years ago

The pH of a solution can be calculated using which formula?

VikaD [51]

Answer:

The formula for calculating pH is pH=−log[H_3O+ ]

pH is the negative logarithm (to base 10) of hydronium ion concentration

The pH Formula can also be expressed as

PH= - log[H+ ]

5 0

2 years ago