For an insoluble compound, how might you prevent formation of the precipitate

Which of the following produces a physical change in the shape of an object?

bogdanovich [222]

Answer:Any change that occurs without altering the chemical composition of a substance is a physical change. Physical changes can include changing the color, shape, state of matter, or volume of a substance. It is crucial to remember that physical changes never alter the chemical makeup.

Explanation:

i hope that helps u try to figure it out a little bit sorry i couldn't find your answer i didn't have much to go off of

8 0

2 years ago

A solution is made by dissolving

barxatty [35]

Answer:

4.52 mol/kg

Explanation:

Given data:

Mass of lithium fluoride = 22.1 g

Mass of water = 188 g

Molality = ?

Solution:

Molality:

It is the number of moles of solute into kilogram of solvent.

Formula:

Molality = number of moles of solute / kilogram solvent

Mathematical expression:

m = n/kg

Now we will convert the grams of LiF into moles.

Number of moles = mass/ molar mass

Number of moles = 22.1 g/ 26 g/mol

Number of moles = 0.85 mol

Now we will convert the g of water into kg.

Mass of water = 188 g× 1kg/1000 g = 0.188 kg

Now we will put the values in formula.

m = 0.85 mol / 0.188 kg

m = 4.52 mol/kg

8 0

2 years ago

What is the compound name to C5H6?

Masja [62]

What you know about rollin' down in the deep?

When your brain goes numb, you can call that mental freeze

When these people talk too much, put that stuff in slow motion, yeah

I feel like an astronaut in the ocean, ayy

3 0

2 years ago

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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

2 years ago

How many grams of water can be produced when 11.7 moles of ethane (C2H6) react with excess oxygen gas?

Elina [12.6K]

Answer: 631.8 g

Explanation:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

It can be seen from the balanced chemical equation, 2 moles of ethane reacts with 7 moles of Oxygen gas to produce 4 moles of carbon dioxide and 6 moles of water.

Ethane is the limiting reagent as it limits the formation of product.

Thus, if 2 moles of ethane produce 6 moles of water.

11.7 moles moles of ethane produce= $\frac{6}{2}\times 11.7=35.1 moles$ of water.

Mass of water= no of moles $\times$ Molar mass

Mass of water= 35.1 $\times$ 18g/mol= 631.8 g

4 0

2 years ago

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