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harkovskaia [24]
3 years ago
14

Atoms that have lost or gained one or more electrons are called ________.

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
3 0
Atoms that have lost or gained one or more electrons are called ions.
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What is the amount of space that a gas takes up called?
trasher [3.6K]
The answer is:Matter
5 0
2 years ago
Which of the following properties would be the same among isotopes of the same element?
Viefleur [7K]

Answer : The properties which would be the same among isotopes of the same element are Flammability and Color

Explanation :

Isotopes are the elements that have same number of electrons and protons but they differ in number of neutrons.

The chemical properties of an element are determined by its valence electrons and since isotope have same number of electrons, they have similar chemical properties.

But isotopes have different number of neutrons. Neutrons are the subatomic particles which are present in the nucleus. They are charge less particles and they have mass. Therefore when the elements have different number of neutrons, their atomic masses differ.

As a result, isotopes have different mass related physical properties.

Let us discuss the given properties now

1) Flammability : Flammability in simple words means the ability of the substance to catch fire or the ability to burn. This is mainly dependent on the reactivity which is in turn dependent on the number of electrons.

Since the isotopes have same number of electrons, they show similar flammability.

2) Density : Density is defined as mass per unit volume. This property involves mass and we know that isotopes have different masses. Therefore they have different densities too.

3) Melting point : Melting point is a physical property which is mainly dependent on the molecular weight of the substance. Since isotopes have different masses, they tend to show different melting points.

4) Color : The color of the element is mainly dependent on the arrangement of electrons around the nucleus. Isotopes have same number of electrons which means the arrangement would also be the same. Therefore the isotopes of the same element do not show different colors.

The properties which would be same among isotopes of the same element are

1) Flammability

4) Color

5 0
3 years ago
Calculate the mole fraction of Ba Cl 2 in an aqueous solution prepared by dissolving 0.400 moles of Ba Cl 2 in 850.0 g of water.
DaniilM [7]

Answer:

0.0084

Explanation:

The mole fraction of BaCl₂ (X) is calculated as follows:

X = moles BaCl₂/total moles of solution

Given:

moles of BaCl₂ = 0.400 moles

mass of water = 850.0 g

We have to convert the mass of water to moles, by using the molecular weight of water (Mw):

Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol

moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol

The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):

total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol

Finally, we calculate the mole fraction:

X = 0.400 mol/47.6 mol = 0.0084

4 0
2 years ago
Hi, does anyone want to be friends ?
Olegator [25]

Answer:

sure sure

Explanation:

5 0
2 years ago
Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -
kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

\Delta H_{C_3H_8}=72.4kJ/mol

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0
2 years ago
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