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3 years ago

Which sentence most accurately describes electrically charged objects?

Chemistry

2 answers:

ivolga24 [154]3 years ago

7 0

Answer:

They are attracted to one other without coming into contact.

Explanation:

In conclusion, an electrically neutral object is an object that has a balance of protons and electrons. In contrast, a charged object has an imbalance of protons and electrons. The type of charge (positive or negative) is determined by whether the protons or the electrons are in excess

Flura [38]3 years ago

4 0

Answer:

A.They are attracted to one other without coming into contact.

Explanation:

Hope that helped! :)

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When comparing the two elements K and Ge , the more metallic element is__________ based on periodic trends alone.

Inessa [10]

Answer:

Option D and Option B

When comparing the two elements K and Ge , the more metallic element is_____K_____ based on periodic trends alone.

When comparing the two elements Sb and Pb , the more metallic element is_____Pb______ based on periodic

Explanation:

The metallic characteristic increases when we move down a column in a periodic table or when we move left in the row.

Potassium and Germanium are located on the same row, but germanium lies on the right side of potassium. Thus, potassium (K) is more metallic than Germanium (Ge)

While Lead (Pb) lies to left of Sb in the adjacent column and is also lies below Sb. Hence Pb is more metallic than Sb

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3 years ago

Which two letters has the most kinetic energy?<br>W<br>X<br>Y<br>Z

hichkok12 [17]

Answer:

x and y

Explanation:

3 0

2 years ago

Read 2 more answers

(a) 4.12 x 10^15 atoms U

DochEvi [55]

Answer:

1.63ₓ10⁻⁶ g of U

139.03 g of H

0.385 g of O

141.8 g of Pb

Explanation:

In first place, we need to convert the number of atoms to moles, as we know that 1 mol of anything occupies 6.02×10²³ particles

Therefore:

4.12×10¹⁵ atoms of U . 1 mol / 6.02×10²³ atoms = 6.84×10⁻⁹ moles of U

8.37×10²⁵ atoms of H . 1 mol /6.02×10²³ atoms = 139.03 moles of H

1.45×10²² atoms of O . 1 mol /6.02×10²³ atoms = 0.0241 moles of O

4.12×10²³ atoms of Pb . 1 mol /6.02×10²³ atoms = 0.684 moles of Pb

Moles . Molar mass = Mass (g)

6.84×10⁻⁹ moles of U . 238.03 g/mol = 1.63ₓ10⁻⁶ g of U

139.03 moles of H . 1 g/mol = 139.03 g of H

0.0241 moles of O . 16 g/mol = 0.385 g of O

0.684 moles of Pb . 207.2 g/mol = 141.8 g of Pb

8 0

3 years ago

Which of the following additions to alkenes occur(s) specifically in an syn fashion?

BlackZzzverrR [31]

Answer:

E) A, B, and C

Explanation:

Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.

Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.

8 0

3 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

3 years ago