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3 years ago

Please answer these questions

2 answers:

7 0

I found number one
Atoms are not drawn to scale. Molecules of compounds have atoms of two or more different elements. For example, water (H2O) has three atoms, two hydrogen (H) atoms and one oxygen (O) atom. Methane (CH4), a common greenhouse gas, has five atoms, one of carbon (C) and four of hydrogen (H, see Fig.

VMariaS [17]3 years ago

3 0

Explanation:

1) Is imposible to determine an element from a simple drawing, more information is need or a higher level of details. Also, a "red blob of atoms" sounds like the description of a molecule (a union of two or more atoms) that may or may not be formed by only one element.

2) There are 92 natural elements, the other ones are synthetic and/or radioactive. You can identify them in the periodic table because their molecular weights are between ( ).

3) False, a molecule is a particular arrangement of atoms that can be from one or more elements. An element is a particular arrangement of atomic particles (neutrons, protons and electrons.)

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A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo

Kruka [31]

Answer: The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

Calculating the molality of solution:

To calculate the mass of solvent, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m

5 0

4 years ago

Stacy runs a paper chromatography experiment using the same ink, but two different solvents.

Pie

Answer:

Did you ever find the answer??

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the energy in joules and calories required to heat 25.0 g of water from 12.5C to 25.7C

Alex Ar [27]

The heat (Q) required to raise the temp of a substance is:Q=m∗Cp∗ΔT where m is the mass of the object (25.0g in this case), Cp is the specific heat capacity of the substance (for water Cp = 1.00cal/gC, or 4.18J/gC,
and Dt is the change in temp.
You'll have to solve this twice, once with the Cp in calories, and once with the Cp in joules.
1380.72 Joules

8 0

3 years ago

write the ground-state electron configurations of the seventh and eighth elements in the first transition series in Universe fif

Tju [1.3M]

Answer: Check explanation.

Explanation:

Transition metals are metallic elements that can be found in the Groups IVB–VIII, IB, and IIB on the periodic chart.

The question require us to write down the GROUND state electronic configuration of the first fifth transition metal, sixth transition metal, seventh transition metal element and the eighth transition metal.

NOTE: we are Starting from Argon, which has 18 electrons.

The fifth transition metal is Manganese,Mn. Manganese has 25 electrons, that is, 25- 18= 7. Therefore, it needs seven electrons to complete the configuration.

Hence, The ground state electronic configuration = [Ar)] 3d5. 4s2.

The first sixth Transition metal is iron,Fe. Iron has 26 electrons, that is, 26 - 18 = 8. Therefore, it need eight Electrons to complete the ground state electronic configuration.

Hence, the ground state electronic configuration of Fe= [Ar] 3d6. 4s2.

The first seventh transition metal is Cobalt, Co. It has 27 Electrons, therefore, 27- 18 = 9. Therefore, it needs 9 Electrons to complete its ground state electronic configuration.

Ground state electronic configuration of Co= [Ar] 3d7. 4s2.

The first eight Transition metal is Nickel. It has 28 electrons. Therefore, 28-18= 10. So, it needs 10 Electrons to complete its ground state electronic configuration.

Hence, the Ground state electronic configuration of Ni= [Ar] 3d8. 4s2.

5 0

3 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

3 0

2 years ago