How many atoms are in sodium

Write the formula for the compound formed between potassium and sulfur. ks ks2 k2s k2so3 k3s2

Mars2501 [29]

The answer should be K2S

6 0

3 years ago

You are given a protein solution with a concentration of 0.15 mg/ml.

Thepotemich [5.8K]

Answer:

We need 2.933 L of 0.15 mg /mL of protein solution.

Explanation:

Concentration of given solution $C_1 = 0.15 mg/mL$

1 mg = 0.001 g , 1 mL = 0.001 L

$C_1=\frac{0.15\times 0.001 mg}{1\times 0.001 L}=0.15 g/L$

Molecular weight of protein = 22,000 Da =22,000 g/mol

Initial concentration in moles/liter:

$C_1=\frac{0.15 g/L}{22,000 g/mol}=6.8182\times 10^{-6} mol/L$

Initial concentration in micromoles/mL :

1 L = 1000 mL

$C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000 mL}=6.8182\times 10^{-3} \mu mole/ mL$

Initial concentration in micromoles/microLiter :

1 L = 1000,000 μL

$C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000000 \mu L}=6.8182\times 10^{-6}\mu mol/\mu L$

Moles of protein required = 20 μmoles

n(Moles)=C(concentration) × V(Volume of solution)

$20 \mu mol=6.8182\times 10^{-6}\mu mol/\mu L\times V$

$V =\frac{20 \mu mol}{6.8182\times 10^{-6}\mu mol/\mu L}$

$V=2.933\times 10^6 \mu L = 2.933 L$

We need 2.933 L of 0.15 mg /mL of protein solution.

6 0

3 years ago

Can someone plz answer my Q.15 for me plz it is do in 5 mins

ladessa [460]

Answer:

i would but there is nothing to awnser soooo!!! sorry

Explanation:

7 0

3 years ago

Read 2 more answers

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

nata0808 [166]

The equilibrium constant for this reaction at 350°C is determined as 5.85.

<h3>Concentration of each component</h3>

concentration of bromine, C(Br) = 0.6 mol/1 = 0.6

concentration of iodine, C(I) = 1.6 mol/1 = 1.6

<h3>Create an ICE table</h3><h3>What is ICE table?</h3>

An ICE table is a tabular system of keeping track of changing concentrations in an equilibrium reaction.

ICE is an abbreviation that stands for initial, change, equilibrium.

Create ICE table for the reactants and products formed;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -1.19 -1.19 1.19

E 0.6 - 1.19 1.6 - 1.19 1.19

E = -0.59 0.41 1.19

<h3>Equilibrium constant </h3>

The equilibrium constant is calculated as follows;

KC = [IBr]²/[Br][I]

KC = (1.19²) / (0.59 x 0.41)

KC = 5.85

Thus, the equilibrium constant for this reaction at 350°C is determined as 5.85.

Learn more about equilibrium constant here: brainly.com/question/19340344

#SPJ1

4 0

2 years ago

When 42.5 g of benzamide (C7H7NO) are dissolved in 500 g of a certain mystery liquid X, the freezing point of the solution is 6.

LuckyWell [14K]

Answer:

21.3g of Fe(NO₃)₃ are required to produce the same depression in freezing point of solution.

Explanation:

Freezing point depression is a colligative property which indicates, in determined mixture, the freezing point of solution is lower that the freezing point of the solvent, according the amount of solute.

Formula is: ΔT = Kf . m . i

Kf is the cryoscopic constant, which is particular for each solvent. We do not have that data, so we need to find it out in order to solve the question:

ΔT = Freezing point of pure solvent - Freezing point of solution.

This data is known → 6.40°C

m, means molality, moles of solute in 1kg of solvent. Let's get the moles of benzamide: 42.5 g . 1mol / 121g = 0.351 moles

m = 0.351 mol / 0.5kg = 0.702 m

As benzamide is an organic compound i, = 1. i are the number of ions dissolved in solution. Let's find out Kf:

6.40°C = Kf . 0.702 m . 1

6.40°C /0.702m = 9.11 °C/m.

Let's go to the next question.

ΔT is the same → 6.40°C

But this is, an inorganic salt, a ionic salt: Fe(NO₃)₃ → 1Fe³⁺ + 3NO₃⁻

For this case, we have 1 mol of Iron(III) and 3 nitrates, so i = 4

Let's replace data: 6.40°C = 9.11 °C/m . m . 4

6.40°C / (9.11 m/°C . 4) = 0.176 m

This data represents that, in 1 kg of solvent we have 0.176 moles of nitrate.

Mass of solvent X required in this case is 0.500 kg, so, the moles that are contained are: 0.500 kg . 0.176 mol/kg = 0.088 mol

Let's determine the mass of salt: 0.088 mol . 241.85g /1mol = 21.3 g

5 0

3 years ago