1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vladimir79 [104]
3 years ago
11

Plssssssssssss help me

Chemistry
1 answer:
Kryger [21]3 years ago
6 0
I can’t see the picture its too small
You might be interested in
Help 4 me???? ples????
frutty [35]

Answer:

its the first one.

Explanation:

If sodium loses an electron, it now has 11 protons, 11 neutrons, and only 10 electrons, leaving it with an overall charge of +1

3 0
3 years ago
Read 2 more answers
Carboplatin (C6H12N2O4Pt) is a platinum-containing compound that is used to treat certain forms of cancer. This compound contain
givi [52]

Answer:

The answer to your question is:  $ 35.6

Explanation:

C₆H₁₂N₂O₄Pt

Platinum = 52.5 %

Price = $1047 / troy ounce

cost of platinum = ? of 2 g

1 troy ounce = 480 grains

1 grain = 64.8 mg

Process

Get 52.5 % of 2 g

                               2 g -----------------  100 %

                               x     -----------------  52.5%

                              x = (52.5 x 2) / 100

                              x = 1.05 g

                               1 g --------------------- 1000 mg

                               1.05 g ----------------   x

                               x = 1050 mg of Pt

                               1 grain ---------------- 64.8 mg

                              x            ---------------  1050 mg

                              x = 16.2 grains

                              480 grains ---------------- 1 troy ounce

                              16.2 grains ----------------   x

                               x = (16.2 x 1) / 480

                                x = 0.034 troy ounce

                              $ 1047 ------------------  1 troy ounce

                                x        -------------------  0.034

                                x = (0,034 x 1047) / 1

                                x = $ 35.6

3 0
3 years ago
What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 321 m
xxMikexx [17]

Answer: The osmotic pressure of a solution is 53.05 atm

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (methanol) = 22.3 g  

Volume of solution = 321 mL

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

\pi=1\times \frac{22.3\times 1000}{32.04\times 321}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K

\pi=53.05atm

Hence, the osmotic pressure of a solution is 53.05 atm

7 0
3 years ago
3. Mr. Hill has 27 students in his class
stich3 [128]
Just dived both numbers by two
4 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
Other questions:
  • Five more than two times the square root of a number is fifteen. What’s the number
    14·2 answers
  • What is the easiest way for atoms without a full valence shell to gain a full valence shell?
    6·2 answers
  • Which Gender Has Better Reflexs
    9·2 answers
  • How many valence electrons does group 18 have
    6·2 answers
  • What is the mass of 8.87 x 10^24 atoms of carbon?<br><br> PLEASE HELP!!
    7·1 answer
  • Use the table to answer the question
    13·2 answers
  • Adaptation: Question 1 Which best explains why a population of a certain species of moose , over five generations , developed fe
    12·1 answer
  • Drag and drop the following building materials from lowest to highest based on their porosity
    5·2 answers
  • Please help me please.
    11·1 answer
  • Calculate the mass of copper that could be made from 4.0g of copper oxide
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!