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Lapatulllka [165]
3 years ago
10

How many moles of H2SO4 are present in 0.500 L of a 0.150 M H2SO4 solution?

Chemistry
1 answer:
bonufazy [111]3 years ago
3 0

Answer: 0.075

Explanation:

(concentration in molarity)(volume in liter) = answer

0.15 mol/L *0.500L = 0.075 mol

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Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct
Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈  +  25/2O₂  →   8CO₂  +  9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8)  = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg  1.12×10¹⁴ g / 1000 =  1.12×10¹¹ kg

6 0
3 years ago
The spectrochemical series is I < Br< < Cl^- < F
vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0
3 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0
3 years ago
How many grams are there in 2.3x10^24 atoms of silver
Sindrei [870]
First calculate using
n=N/Na (Na is avagardros number)

then substitute n into:
n=m/M (M is the molar mass)

your answer is: 412.06 g
7 0
3 years ago
You have to decide to change the temperature at which you run a certain reaction in hope of obtaining more product more quickly.
Alexandra [31]

Answer:

see explanation

Explanation:

The reaction has a negative rate law; i.e., Rate = - ΔConcentration / ΔTime which is graphically a negative slope for the plot of Rate as a function of reactant concentration. => Rate ∝ f(Reactant Concentration). However, by raising the temperature, an increase the probability of reaction occurs and the formation of more product.

4 0
3 years ago
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