Which of the following will decrease the rate of dissolving?*

Which objects cannot be observed in detail without a microscope?

Sonbull [250]

The objects which cannot be observed in detail without a microscope

include the following:

Red blood cell

Bacterium

<h3>What is a Microscope?</h3>

A microscope is an instrument which is used to view smaller objects such as

microbe,cells, tissues etc. This instrument is used in viewing the different

cells found in the body as they can't be seen with the eye.

The remaining options which can be seen with the eyes don't require

the use of microscopes.

Read more about Microscope here brainly.com/question/25268499

3 0

2 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

When iridium-192 is used in cancer treatment, a small cylindrical piece of 192ir, 0.6 mm in diameter and 3.5 mm long, is surgica

Ivan

Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?

Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.

V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3

Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":

9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)

= 6.96 x 10^19 atoms of Ir-122 are present.

4 0

3 years ago

PLEASE HELP ASAP !!

Ivenika [448]

Answer:

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Theoretical yield of H₂SO₄ = 213 g

percent yield of H₂SO₄ = 94 %

Explanation:

Data Given:

volume of SO₂ = 48.6 L

mass of H₂SO₄ = 200 g

balance equation = ?

theoretical yield = ?

percent yield = ?

Solution:

Part 1:

first we have to write a balance equation for the reaction

SO₂ gas react with water (H₂O) and excess oxygen

The balanced equation is as under

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

Part 2:

Now we have to find theoretical yield

First look at the balance reaction

2SO₂ + O₂ + 2H₂O -----------> 2H₂SO₄

2 mol 2 mol

2 moles of SO₂ give gives 2 moles of H₂SO₄

Now calculate volume of 2 moles of SO₂ and mass of 2 moles of H₂SO₄

volume of 2 moles of SO₂

Formula used

volume of gas = no. of moles x molar volume . . . . . . (1)

molar volume of SO₂= 22.4 L/mol

Put values in above formula (1)

volume of gas = 2 mol x 22.4 L/mol

volume of gas = 44.8 L

volume of 2 mole of SO₂ = 44.8 L

Now,

Find mass of 2 mole H₂SO₄

Formula Used

mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of H₂SO₄ = 2 (1) + 32 + 4(16)

molar mass of H₂SO₄ = 98 g/mol

put values in equation 2

mass in grams = 2 mol x 98 g/mol

mass in grams = 196 g

mass of 2 mole of H₂SO₄ = 196 g

** So,

Now we come to know that

44.8 L of SO₂ gives 196 g of H₂SO₄ then how many grams of the H₂SO₄ will be produced by 48.6 L of SO₂

Apply unity Formula

44.8 L of SO₂ ≅ 196 g of H₂SO₄

48.6 L of SO₂ ≅ X g of H₂SO₄

Do cross multiplication

g of H₂SO₄ = 196 g x 48.6 L / 44.8 L

g of H₂SO₄ = 213 g

So that is why the theoretical yield of H₂SO₄ is 213 g

Theoretical yield of H₂SO₄ = 213 g

Part 3

Calculate Percent Yield:

Formula used for this purpose:

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 200 g/ 213 g x 100 %

percent yield = 94 %

So percent yield of H₂SO₄ = 94 %

8 0

3 years ago

Consider the hydrate FeSO4 • 7H2O.

DochEvi [55]

The molar mass of the hydrate is 278.06
The molar mass of the anhydrous salt is 151.92
The molar mass of water in the hydrate is 126.14

7 0

3 years ago

Read 2 more answers