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adell [148]
3 years ago
12

What is the correct electron configuration for an element with 5 electrons in

Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

it is c

Explanation:

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A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the t
Mariulka [41]

Answer:

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

<em> Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>

Explanation:

Using the formula of heat, Q = mc∆T  

where Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron  = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C  = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

5 0
2 years ago
Mr. Chem S. Tree added water to 250. ML of a 2.50 M NaOH solution, until the final volume was 500. ML. What is the new molarity
tresset_1 [31]

Answer:

molarity of diluted solution = 1.25 M

Explanation:

Using,          

C1V1 (Stock solution) = C2V2 (dilute solution)

given that

C1 = 2.50M

V1 = 250ML

C2 = ?

V2 = 500ML

2.50 M x 250 mL = C2 x 500 mL

C2 = (2.50 M x 250 mL) / 500 mL

C2 = 1.25 M

Hence, molarity of diluted solution = 1.25 M

4 0
3 years ago
Balancing Al+ Agcl, i have problems with this so help!!!
EleoNora [17]

Answer:

Al  + 3AgCl  →  AlCl₃  + 3Ag

Explanation:

The given equation is:  

      Al  + AgCl  →  

We are to find the product and hence balance the equation. This problem is a simple single replacement reaction.

By virtue of this, Aluminum will displace Ag from the solution:

                       Al  + AgCl  →  AlCl₃  + Ag

    We then balance the equation:

                        Al  + 3AgCl  →  AlCl₃  + 3Ag

                                             

5 0
3 years ago
What is the mass percent of oxygen (0) in SO2?
vazorg [7]

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

5 0
3 years ago
Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.
MrRissso [65]
<h2>Answers to the rest of the assignment:</h2>

**Check for proof photos at the bottom.**

__________________________________________________________

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.  

Cranberry juice at the store

A. homogeneous

C. solution

Smoke  

B. heterogeneous  

D. colloid

__________________________________________________________

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.  

Blood

B. heterogeneous  

E. suspension

Salad dressing

B. heterogeneous  

E. suspension

__________________________________________________________

Shown here is a person shaving.  Under magnification, the shaving foam might look like the image above the shaver. What type of mixture does the foam demonstrate? Give your reasoning.

Foam is a colloid. Colloids includes gas dispersed in a liquid and it also includes gas dispersed in a solid

On the next slide, you can select any answers you want, or you can select nothing. There's no wrong answer.  

__________________________________________________________

A student squeezes several oranges to make a glass of orange juice. The juice contains pieces of orange pulp mixed with the juice. Explain why this drink can be considered a combination of a suspension and a solution.

The juice contains sugars, plant pigments, and other chemicals dissolved in water. This is a solution. The pieces of orange pulp will rise to the top or settle to the bottom of the juice if it is allowed to sit. The pieces of pulp mixed with the juice form a suspension

On the right side, you can also select any answers or none at all. There is no wrong answer.

__________________________________________________________

<h2>Explanation:</h2>

There are two main types of mixtures, homogeneous mixtures and heterogeneous mixtures. The components of a homogeneous mixture are evenly distributed throughout the mixture. The properties of a mixture are the same everywhere. The components of a heterogeneous mixture are not evenly distributed. Different regions of this mixture have different   properties.

Particles in a homogeneous mixture do not settle down or separate when left alone. Particles in a heterogeneous mixture eventually separate or settle when left alone.

  • Colloids are a type of heterogeneous mixture that do not naturally settle out quickly, and can be separated with different methods.
  • A suspension is another type of heterogeneous mixtures in which the components of the mixture will quickly settle out or can be filtered or separated.

Here are photos of Edge just incase.

6 0
3 years ago
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