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Nezavi [6.7K]
3 years ago
11

If an element is composed of atoms with an atomic number of 6 and a mass number of 14 then a neutral atom of this element contai

ns?
Chemistry
1 answer:
kvasek [131]3 years ago
8 0

Answer:

Proton 6

Neutron 8

Electron 6

Explanation:

A neutral atom of an element is one that is not charged. There are 2 deciders if the neutrality of an atom. These are the electrons and the protons. Since the electrons are positively charged leading to a net positive core(nucleus), there must be a mechanism that supplies same amount of negative charges to maintain the electrical neutrality of the atom.

Since they both carry similar albeit opposite charges, the number of electrons would be equal to the number of protons. This is essentially why an atom is adjudged to be electrically neutral.

Now to get the number of neutrons in an atom, we simply subtract the proton number from the mass number.

In this case specifically, the number of neutrons is 14 - 6 = 8

Since it has also been said that the atom is neutral, the number of electrons too will be 6 electrons

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To convert from liters/second to cubic gallons/minute, multiply the number of liters/second by 15.850 0 0.0353 00.2642 0 60
ivanzaharov [21]

Answer: 15.850

Explanation:

The conversion used from liters to gallons is:

1 L = 0.264172 gallon

The conversion used from sec to min is:

60 sec = 1 min

1 sec =\frac{1}{60}\times 1=0.017min

We are asked: liters/sec = gallons/min

liters/sec=\frac{0.264172}{0.017}=15.850gallons/min

Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.

8 0
4 years ago
Which is the correctly balanced chemical equation for the reaction of KOH and H2SO4?
Ksenya-84 [330]

Answer:

3) 2KOH + H2SO4 → K2SO4 + 2H2O

Explanation:

The balanced chemical equation for the reaction of KOH and H2SO4 is,

→ 2KOH + H2SO4 → K2SO4 + 2H2O

Hence, option (3) is the correct answer.

5 0
3 years ago
Determine how many moles of O2 are required to react completely with 4.6 mol of C4H10
kondaur [170]

Answer:

29.9 moles

2C²H¹⁰ needs 13 moles of O²

4.6 C⁴H¹⁰ needs X moles of O²

X= 13× 4.6 ÷ 2 = 59.8 ÷ 2 = 29.9 moles

3 0
3 years ago
How many grams are in 8 moles of AI?
ella [17]

Answer:

c. 216 g

Explanation:

There is actually 215.852304 but when you round it, you will get 216

4 0
3 years ago
A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe
nikitadnepr [17]

Answer:

\large \boxed{34.2\, ^{\circ}\text{C}}

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:  

For the metal:

m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

For the water:

m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}

\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}

\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

3 0
3 years ago
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