All categories

3 years ago

PLEASEEE HELPPPP ASAP!!

Chemistry

2 answers:

Goshia [24]3 years ago

8 0

Atoms are made up of three subatomic particles called protons, neutrons, and electrons.

Protons and neutrons are located in the nucleus.

All protons have a positive charge.

All neutrons have no charge or are neutral.

Electrons orbit around the nucleus and have a negative charge.

Firlakuza [10]3 years ago

4 0

electrons

protons

neutrons

nucleus

positive

neutral

negative

You might be interested in

What element is found in group 5 period 3?

tatuchka [14]

Answer:

Phosphorus

Explanation:

A group is a vertical column on the periodic table.

A period is a horizontal row on the periodic table.

5 0

2 years ago

Which statement correctly compares sound and light waves?

Mrac [35]

Answer:

Light waves carry energy parallel to the motion of the wave, while sound waves carry energy perpendicu

Explanation:

5 0

3 years ago

Does protecting animals so they won’t go extinct help with climate change?

Alchen [17]

Answer:

Yes

Explanation: Yes it depends on what kind of animal but yes it can help

6 0

2 years ago

Read 2 more answers

Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V

malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

4 0

2 years ago

A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem

BartSMP [9]

Answer:

$Q_{metal} = -6799\,J$

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

$Q_{metal} = - Q_{w}$

$Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})$

$Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)$

$Q_{metal} = -6799\,J$

6 0

3 years ago

Read 2 more answers