Consider the following reaction:

What happens in the process of beta decay?

Andrej [43]

Answer:

A neutron transforms into a proton and an electron.

Explanation:

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6 0

3 years ago

The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by

maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:

p₁ = χ₁p₁°

where

χ₁ = the mole fraction of the solvent

p₁ and p₁° are the vapour pressures of the solution and of the pure solvent

The formula for vapour pressure lowering Δp is

Δp = p₁° - p₁

Δp = p₁° - χ₁p₁° = p₁°(1 – χ₁) = χ₂p₁°

where χ₂ is the mole fraction of the solute.

Step 1. Calculate the mole fraction of glucose

n₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu

n₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O

χ₂ = n₂/(n₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62

Step 2. Calculate the vapour pressure lowering

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr

Step 3. Calculate the vapour pressure

p₁ = p₁° - Δp = 23.8 torr – 0.4430 torr = 23.4 torr

3 0

3 years ago

A 0. 462 g sample of a monoprotic acid is dissolved in water and titrated with 0. 180 m koh. what is the molar mass of the acid

Len [333]

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

<h3>For the calculation of molarity of solution</h3>

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

<h3>Chemical equation for the reaction</h3>

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

<h3>To calculate the molar mass for given number of moles</h3>

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

learn more about molar mass or moles:

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3 0

1 year ago

7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.

beks73 [17]

Answer:

$\Delta H_{f,C_3H_4}=276.8kJ/mol$

Explanation:

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In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

$\Delta H_{rxn} =- m_wC_w\Delta T$

We plug in the mass of water, temperature change and specific heat to obtain:

$\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ$

Now, this enthalpy of reaction corresponds to the combustion of propyne:

$C_3H_4+4O_2\rightarrow 3CO_2+2H_2O$

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

$\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}$

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

$\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol$

Now, we solve for the enthalpy of formation of C3H4 as shown below:

$\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}$

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

$\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol$

Best regards!

7 0

3 years ago

How did gus get ao gay ??

Mumz [18]

Magic, that how

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4 0

3 years ago