All categories

3 years ago

Two atoms that are isotopes of one another must have the same number of what?

Chemistry

2 answers:

erastovalidia [21]3 years ago

6 0

They must have the same number of protons

Alex3 years ago

4 0

Answer:

Protons!

Explanation:

You might be interested in

What is the coefficient of diphosphorus trioxide? (p2o3)

denis-greek [22]

The cofficent is 2 for diphosphorus trioxide

7 0

3 years ago

Questions and Problems 22 When water is heated, the temperature eventually reaches a constant value and forms a plateau on the g

Softa [21]

Answer:

It indicates phase changes:<em> melting then later boiling.</em>

Explanation:

On the heating curve of water, the plateau marks phase change transitions.The temperature remains constant during this period.The first phase change is melting of water.The second phase change is boiling of water to gas.

6 0

3 years ago

Solve the following 2 questions below, involving elements and reactions. *

padilas [110]

Answer:

1. 40 grams of NaCI

2. 20 grams of Barium Peroxide

Explanation:

<h3><u>In this question, we have to divide to find the amount of each element in a drop. </u></h3>

For Question 1, there are 400 grams of NaCI. There are 10 drops.

So, we have to divide 400 by 10.

400 / 10 = 40

We get 40 as our result, so 40 is the answer.

<h3><u>Same for this question, we use the process we did before.</u></h3>

There are 800 grams of barium peroxide, and there are 40 drops.

So, we divide 800 by 40.

800 / 40 = 20

We have the answer as 20.

Our final answers are going to be listed below:

Question 1 - 40
Question 2 - 20

Hope this helps you!

5 0

3 years ago

Read 2 more answers

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.220 M sodium benzoate

ddd [48]

Answer:

34.5 mL

Explanation:

Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .

[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20

First we need to calculate the ratio of the conjugate base and acid

We know, Henderson Hasselbalch equation

pH = pKa + log [C6H5COO-] /[C6H5COOH]

4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]

log [C6H5COO-] /[C6H5COOH] = 4.00-4.20

= - 0.20

Antilog from both side

[C6H5COO-] /[C6H5COOH] = 0.631

Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume

Sum of volume = 100 mL = 0.100 L

volume of benzoic acid = x

volume of benzoate = 0.10 -x ,

so Volume of acid + volume of conjugate base = 0.100 L

[C6H5COO-] /[C6H5COOH] = 0.631

[C6H5COO-] = 0.631 * [C6H5COOH]

0.120 (0.1-x) = 0.631 *0.100x

So, x = 0.065

So, volume of benzoic acid = x = 0.0655 L

= 65.5 mL

So, volume of sodium benzoate = 0.1 -x

= 0.1-0.0655

= 0.0345 L

= 34.5 mL

So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.

5 0

3 years ago

Read 2 more answers

Kinetic Energy Problems = KE = 0.5m x 12

Brut [27]

Explanation:

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

1. Mass, m = 5 kg

Speed, v = 2 m/s

$K_1=\dfrac{1}{2}\times 5\times (2)^2\\\\=10\ J$

2. Mass, m = 71 kg

Speed, v = 1 m/s

$K_2=\dfrac{1}{2}\times 71\times (1)^2\\\\=35.5\ J$

3. Mass, m = 71 kg

Speed, v = 5 m/s

$K_3=\dfrac{1}{2}\times 71\times (5)^2\\\\=887.5\ J$

4. Mass, m = 1816 kg

Speed, v = 25 m/s

$K_4=\dfrac{1}{2}\times 1816\times (25)^2\\\\=567500\ J$

8 0

3 years ago