What is the mass of 6.02 x 1023 particles of rubidium carbonate

What is the molarity when 0.181 moles of MgNO3)2 are dissolved in enough water to make 0.750 L solution?

eimsori [14]

Answer:

Molarity = 0.24 M

Explanation:

Given data:

Number of moles of Mg(NO₃)₂ = 0.181 mol

Volume of solution = 0.750 L

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

by putting values,

Molarity = 0.181 mol / 0.750 L

Molarity = 0.24 M

4 0

3 years ago

What is the air temperature 8,850 m in the air?

evablogger [386]

The temperature is gonna be

5 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

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

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

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

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

3 years ago

The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O

Mila [183]

Answer:

41.3kJ of heat is absorbed

Explanation:

Based in the reaction:

Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ

<em>1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.</em>

63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:

63.4g Fe₃O₄ × (1mol / 231.533g) = <em>0.274moles of Fe₃O₄</em>

These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:

0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =

<h3>41.3kJ of heat is absorbed</h3>

<em />

6 0

3 years ago

The Brahman cattle have good resistance to high temperature, but its meat is poor and tough to eat. The English shorthorn cattle

N76 [4]

Answer:

d

Explanation:

4 0

3 years ago

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