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4 years ago

Why is ice in a glacier considered to be a mineral, but water from a glacier is not?

1 answer:

5 0

Minerals need to be solid with a crystalline structure. Ice has these properties, liquid water does not. (i hope that helps you)

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In a redox reaction, what role does the reducing agent play?

viktelen [127]

So the oxidizing agent will receive electrons from the reducing agent and the oxidation agent will take electrons from the reducing agent.

3 0

3 years ago

Read 2 more answers

What is the mass of 3.8 × 10^24 atoms of argon (ar)? 0.0039 g 0.16 g 6.3 g 250 g

ale4655 [162]

Atomic mass Ar => 39.948 a.m.u

39.948 g --------------- 6.02x10²³ atoms
?? g -------------------- 3.8x10²⁴ atoms

(3.8x10²⁴) x 39.948 / 6.02x10²³ => 250 g

hope this helps!

5 0

3 years ago

Read 2 more answers

Please help me! Will mark brainliest! Thank you!

Kamila [148]

Answer:

gas

Explanation:

8 0

3 years ago

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

4 0

3 years ago

A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a

Andrew [12]

Answer:

Question 1: 1 s after the motion starts

Question 2: 0 (just when the motion starts)

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

1. From the speedometer shown at the right.

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

Free fall equation: Vf = Vo - gt

Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

2. Time when the speedometer displays a reading of 20 m/s

First, calculate the launching speed:

Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

0 = Vo - gt

Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

20 m/s = 30 m/s - 10 m/s² × t

t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

3. Time when the speedometer displays a reading of 30 m/s

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0

3 years ago