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3 years ago

Classify the following as fluids or solids: warm butter, liquid nitrogen, paper, neon gas, ice.

Chemistry

1 answer:

gogolik [260]3 years ago

6 0

Warm butter- liquid
Liquid nitrogen- gas
paper- solid
neon gas- gas
ice- solid

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Determine the mass of oxygen necessary to react with 2.3 moles of HCl

Orlov [11]

4HCl + O₂ → 2Cl₂ + 2H₂O

mole ratio of HCl : O₂ is 4 : 1

∴ if moles of HCl = 2.3 mol

then mol of O₂ = 2.3 mol ÷ 4
= 0.575 mol

mass of O₂ = moles of O₂ × molar mass of O₂
= 0.575 mol × (16 × 2) g/mol
= 18.4 g

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3 years ago

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Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.

nordsb [41]

<u>Answer:</u> The molality of magnesium chloride is 1.58 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (magnesium chloride) = 75.0

$M_{solute}$ = Molar mass of solute (magnesium chloride) = 95.21 g/mol

$W_{solvent}$ = Mass of solvent = 500.0 g

Putting values in above equation, we get:

$\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m$

Hence, the molality of magnesium chloride is 1.58 m

5 0

3 years ago

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Dvinal [7]

Explanation:

The lack of a control group

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Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

vlada-n [284]

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to $Zn^2^+$ and $Se^2^-$ . Following reaction represents the ionization of ZnSe in solution -

$ZnSe$ ⇄ $Zn^2^+ + Se^2^-$

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have $Na^+$ and $Cl^-$ in the solution which will be formed by the ionization of NaCl.

Now, $Zn^2^+$ in the solution will react with two $Cl^-$ ions to form $ZnCl_2$ as follows -

$Zn^2^++2Cl^-$ ⇄ $ZnCl_2$

Due to this reaction the concentration of $Zn^2^+$ will decrease in the solution and more ZnSe can be soluble in the solution.

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3 years ago