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3 years ago

The relatively high boiling point of water is due to water having

1 answer:

3 0

The answer should be hydrogen bonding. Water only has oxygen and hydrogen in it, which are both nonmetals, so you know the answer cannot be metallic or ionic. It also cannot be nonpolar because the electronegativity of the oxygens will make the molecule polar. You can also know it is hydrogen bonding because it can only take place when a hydrogen is attached to an oxygen, fluorine, or nitrogen. These bonds are very strong attractions, so the molecules are extremely hard to pull apart, creating a high boiling point. Hope that helps!

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If the earth is about 13000 light years away from the outermost edge of the milky way galaxy, what is the total diameter of the

xeze [42]

Answer: The Milky Way is a barred spiral galaxy with a diameter between 150,000 and 200,000 light-years (ly).

3 0

4 years ago

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

6 0

3 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago

A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

3 years ago

True or false weight is affected by gravity

bazaltina [42]

true, in a stronger gravitational field the weight will be larger.

6 0

4 years ago

Read 2 more answers