Plssssssssssss help me

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

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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

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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