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2 years ago

Why does increasing concentration usually increase reaction rate?

Chemistry

2 answers:

nignag [31]2 years ago

8 0

C) There are more particle collision

Alex Ar [27]2 years ago

3 0

The correct answer is C) There are more particle collision

With more particle collision, more reactions are created.

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This electron cloud model of aluminum has _______ energy levels with _______ electrons on its outer energy level. A) 1, 2 B) 2,

Talja [164]

Answer:

Explanation:

Have anice day

7 0

2 years ago

An inflatable toy starts with 1.05 moles of air and a volume of 5.17 liters. When fully inflated, the volume is 8.00 liters. If

padilas [110]

Answer:- 1.62 moles

Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

from given data, $V_1$ = 5.17 L, $n_1$ = 1.05 moles

$V_2$ = 8.00 L, $n_2$ = ?

Let's plug in the values in the formula:

$\frac{5.17L}{8.00L}=\frac{1.05moles}{n_2}$

On cross multiply:

$n_2=\frac{8.00L(1.05moles)}{5.17L}$

$n_2$ = 1.62 moles

So, now the toy contains 1.62 moles of the air.

5 0

3 years ago

If the edge length of the unit cell is 705.2 pm, what is the density of KI in g/cm3.

Kipish [7]

The density is 3.144 g / cm^3.

Explanation:

If effective number of atom in NaCl type structure, z = 4

a = 705.2 pm ⇒ In centimeter = 705.2 $\times$ 10^-10

Na = 6.023 $\times$ 10^23

density = (molecular weight) (z) / (Na) (a^3)

where molecular weight of KI is 166 g,

Z represents the atomic number

density = (molecular weight) (z) / (Na) (a^3)

= (166 $\times$ 4) / (6.023 $\times$ 10^23) $\times$ (705.2 $\times$ 10^-10)

density = 3.144 g / cm^3.

5 0

2 years ago

Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to

AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength $6.32x10^{-36}$ m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

$\lambda =\frac{h}{m*v}$

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

$\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m$

Best regards!

7 0

2 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the conversion factor as 135.58 kJ/ 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

3 years ago