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3 years ago

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A 50.0-g sample of liquid water at 25.0 Â°c is mixed with 23.0 g of water at 79.0 Â°c. the final temperature of the water is ___

_____ Â°c.

1 answer:

I am Lyosha [343]3 years ago

8 0

<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>

<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>

<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>

<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>

<span>73.0<span>Tf</span>=2561 °C</span>

<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>

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How many atoms are in c6h4c12

Mekhanik [1.2K]

Answer:

their are a it is in the chemicle

Explanation:18 carbon and 4 hydrogen

so it is a toatle of 22 atoms

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3 years ago

What is meant by the term neutral in relation to atoms?

Sonbull [250]

Answer:

it means positive and negative charges are equal.

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

when a neutral atom loses a electron positive charge is created because number of protons are positive charge becomes greater than negative charge.

X → X⁺ + e⁻

When a neutral atom accept the electron negative charge is created because negative charge is become greater than positive charge.

X + e⁻ → X⁻

3 0

3 years ago

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

$Molecular\ weight = \frac{Mass\ in\ g}{Moles}$

Molecular weight = $\frac{1.450}{0.00389} = 372.13\ g/mol$

4 0

3 years ago

PLEASE HELP ME ASAPPPP

sattari [20]

Boyle’s law gives the relationship between pressure and volume of gases. It states that at constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
Where P is pressure V is volume and k is constant
P1V1 = P2V2
Parameters at STP are on the left side and parameters for the second instance are on the right side of the equation
P1 - standard pressure - 1.0 atm
Substituting the values in the equation
1.0 atm x 5.00 L = P x 15.0 L
P = 0.33 atm
New pressure is 0.33 atm

5 0

3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago