0.7 moles of uranium to grams of uranium

Convert 150 g/L to the unit g/mL. 15,000 g/mL 15 g/mL 0.15 g/mL 0.0015 g/mL

fredd [130]

1 g/L ------- 0.001 g/mL
150 g/L ----- ?

150 x 0.001 / 1

= 0.15 g/mL

Answer C

6 0

4 years ago

A spinner is divided into sections of equal size, of which some are red, some are blue, and the remaining are green. The probabi

Mariana [72]

There's a total of 3 colors, the percent will add up to 100%

100%-50%(red)= 50%
50%-10%(blue)=40%

There is a 40% chance it will land on green

7 0

3 years ago

Read 2 more answers

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the weighted average score will be 84.1.

Find more about percents here:

brainly.com/question/255442?referrer=searchResults

brainly.com/question/22444616?referrer=searchResults

I hope it helps you!

4 0

3 years ago

Why are some voltaic cells called dry cells even though their chemistry involves water

alexandr402 [8]

They are called dry cells because the electrolyte is a paste.
Hope this helps!
Please give Brainliest!

6 0

2 years ago

Read 2 more answers

For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the

emmainna [20.7K]

Answer:

For 1: The correct answer is Ge.

For 2: The correct answer is Te.

For 3: The correct answer is Ba.

For 4: The correct answer is Ag.

Explanation:

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as $E_i$

$X(g)\rightarrow X^+(g)+1e^-;E_i$

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

Option 1:

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

Option 2:

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

Option 3:

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

Option 4:

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

5 0

3 years ago

0.7 moles of uranium to grams of uranium​

0.7 moles of uranium to grams of uranium