What type of bond would form between two atoms of selenium?

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),

lbvjy [14]

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

5 0

2 years ago

How many capsules containing 75mg of Tamiflu could be produced from 155g of star anise.?

jeyben [28]

From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication = $\frac{(1 x 155)}{2.6}$ = 59 capsules

7 0

2 years ago

A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters

nadya68 [22]

n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9 First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4. We know that 1dm3=1L, so H2SO4's molarity is C=nV=18.0moles1.0L=18M In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so 18.0moles1Lâ‹…98.0g1mole=1764g1L Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution 98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolutionâ†’ masssolution=1764gâ‹…100.0g98g=1800g Therefore, 1L of 98wt% H2SO4 solution will have a density of Ď=mV=1800g1.0â‹…103mL=1.8gmL H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that 100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4 100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O So, H2SO4's mole fraction is molefractionH2SO4=11+0.11=0.9

5 0

2 years ago

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

7 0

2 years ago

What substance is used in the industrial preparation of methyl diantilis to reduce the aldehyde group of 3-ethoxy-4-hydroxybenza

GrogVix [38]

Answer:

Sodium Borohydride (NaBH₄)

Explanation:

Methyl diantilis (2-Ethoxy-4-(methoxymethyl)phenol) is a fragrance compound which smells like Vanilla. This compound is being synthesized from 3-ethoxy-4-hydroxybenzaldehyde also known as Ethyl Vanillin in two steps.

Step 1: Reduction of Aldehydic Group on Ethyl Vanillin:

The benzaldehyde derivative is treated with a mild reducing agent i.e. NaBH₄ (Sodium Borohydride). NaBH₄ is a source of Hydride (H⁻) ion and undergoes nucleophilic substitution reaction yielding 2-ethoxy-4-(hydroxymethyl)phenol.

Step 2: Etherification of 2-ethoxy-4-(hydroxymethyl)phenol:

In the second step 2-ethoxy-4-(hydroxymethyl)phenol is treated with Methanol in the presence of strong acidic polymeric resin known as Amberlyst-15-wet resulting in the formation of Methyl diantilis as shown in attached figure.

5 0

2 years ago