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3 years ago

10

In a reaction that occurs in solution, the volume will not change. What happens to the concentration of the reactants? What happ

ens to the reaction rate as the reaction proceeds?
(will give brainllest)

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

the reaction will increase

Explanation:

You might be interested in

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

<u>Answer:</u> The molecular formula of the compound is $C_4H_{10}O_4$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

<u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.

Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>

Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
Initial temperature (T₁) = –25 °C
Final temperature (T₂) = 0 °
Change in temperature (ΔT) = 0 – (–38) = 38 °C
Specific heat capacity (C) = 2050 J/(kg·°C)
Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 0 °C
Final temperature (T₂) = 100 °C
Change in temperature (ΔT) = 100 – 0 = 100 °C
Specific heat capacity (C) = 4180 J/(kg·°C)
Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 100 °C
Final temperature (T₂) = 160 °C
Change in temperature (ΔT) = 160 – 100 = 60 °C
Specific heat capacity (C) = 1996 J/(kg·°C)
Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>

Heat for –38 °C to 0°C (Q₁) = 31160 J
Heat for melting (Q₂) = 133600 J
Heat for 0 °C to 100 °C (Q₃) = 167200 J
Heat for vaporization (Q₄) = 904000 J
Heat for 100 °C to 160 °C (Q₅) = 47904 J
Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

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7 0

2 years ago

What characteristics of a metallic bond explains some of the properties of metals

d1i1m1o1n [39]

Answer:

See Explanation

Explanation:

Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.

The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.

Due to the nature of the bonding between metal atoms,metals are malleable and ductile.

Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.

8 0

3 years ago

Suppose a mercury thermometer contains 3.250g of mercury and has a capillary that is 0.180mm in diameter.. How far does the merc

Helga [31]

1) You need to get volume of both temperatures by using first attached formula V= Mass/Density $V_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3
At 25 degrees C (V_2e have:

V_2 = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3$

2) Using the second formula you get the height of 0 degree
$h = V/(pi * r^2)


h_1 =(.2390 cm^3)/(pi*(.009 cm)^2) = 939.2 cm$

(radius in cm is $(0.180 mm) / 2 * (1 cm)/(10 mm) or .009 cm$

3) Then with h1 you can easily get the height of 25 degrees
$h_2 =(.2401 cm^3)/(pi*(.009 cm)^2) = 943.5 cm$

Subtract 943.5 cm - 939.2 cm, and obtain a rise in mercury height of 4.3 cm

5 0

3 years ago

Read 2 more answers

6) Which of these elements is a metal?<br> E) C<br> D) N<br> A) O<br> C) Ne<br> B) Ca

sertanlavr [38]

Answer:

Ca

Explanation:

It is on the left side of the periodic table where metals are located.

4 0

3 years ago

Read 2 more answers