The soil of an agricultural field is too basic what can we do to correct it

cyclopropane is a commonly used anesthetic. If a 2.00 L flask contains 3.11 g of cyclopropane gas at 684 mmHg pressure and 23.0

jeka94

Answer:

42.0g/mol

Explanation:

The steps for this question is to use the ideal gas law, and then use n=m/M to find molar mass.

PV = nRT

n= PV/RT

Change 23 degrees to 296.15 K and 684mmHg to kpa

n = (91.1925kpa)(2.00L)/(8.314Lkpa/mol K)(296.15K)

n= 0.074mol

n = m/M

M = m/n

M = 41.98489g/mol

8 0

2 years ago

A substance that speeds up a chemical reaction without being part of the chemical change, itself:

allochka39001 [22]

It is known as a catalyst

6 0

3 years ago

How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 88.1 molesmoles of water vapor?

irga5000 [103]

Answer:

88.1 moles of H₂

Explanation:

Let's make the reaction:

H₂(g) + CO₂(g) → H₂O(g) + CO(g)

Ratio is 1:1

Therefore, 1 mol of hydrogen reacts with 1 mol of carbon dioxide to produce 1 mol of water and 1 mol of carbon monoxide.

In conclusion, 88.1 moles of water must be produced by 88.1 moles of H₂

The reaction can also be written as an equilbrium

H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g) Kc

8 0

4 years ago

In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial te

Zigmanuir [339]

Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol$

$moles=Molarity\times Volume (L)$

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

$n=0.50 M\times 0.150 L=0.075 mol$

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

$n'=1.00 M\times 0.050 L=0.050 mol$

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

$-56 kJ/mol=-\frac{Q}{n}$

$Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J$

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

$q=mc\times (T_{final}-T_{initial})$

where,

q = heat gained = 2800 J

c = specific heat of solution = $4.184 J/^oC$

$T_{final}$ = final temperature = $?$

$T_{initial}$ = initial temperature = $48.2^oC$

Now put all the given values in the above formula, we get:

$2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC$

$T_{final}= 51.54^oC$

51.54°C the final temperature of the calorimeter contents.

7 0

3 years ago

Ion-dipole interactions can occur between any ion and any molecule with a dipole. Identify all of the following pairs of species

Zigmanuir [339]

An ion-dipole interaction is the result of an electrostatic interaction between a charged ion and a molecule that has a dipole. It is an attractive force that is commonly found in solutions, especially ionic compounds dissolved in polar liquids. A cation can attract the partially negative end of a neutral polar molecule, while an anion attracts the positive end of a polar molecule. Ion-dipole attractions become stronger as the charge on the ion increases or as the magnitude of the dipole of the polar molecule increases.

This force of attraction is between an ion and a charge , it is weaker force than covalent bond and ionic bond . EX - The ion dipole interaction takes place between water and sodium ion , in it there is a small charge on oxygen molecule in water which is attracted by sodium charge .

Most commonly found in solutions. Especially important for solutions of ionic compounds in polar liquids.

A positive ion (cation) attracts the partially negative end of a neutral polar molecule.

to learn more about dipole interactions:-

https://brainly.in/question/1157107

4 0

2 years ago