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Lostsunrise [7]

3 years ago

8

Imagine that you have a piece of Styrofoam. Suppose you squeezed it hard enough to change its shape. Briefly describe how this c

ompression would affect its mass, volume, and density.

1 answer:

AleksAgata [21]3 years ago

6 0

Its mass would stay the same, it’s volume would become smaller, and it would become more dense.

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Which would lower the reaction rate?

saveliy_v [14]

Answer:

D

Explanation: Decreasing the temperature of the system would lower the reaction rate.

4 0

3 years ago

A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10

lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0

3 years ago

levacccp [35]

Answer:

A pure substance has a definite chemical composition

4 0

3 years ago

If 4.80 mol Ca mixed with 2 mol N2, which is the limiting reactant? 3Ca (s) + N2 (g) Ca3N2 (s)

Andreyy89

Nitrogen in the limiting reactant x

4 0

2 years ago

Winter is coming and it's time to make sure you have the right amount of antifreeze

ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

molality

$\tt =\dfrac{107.608}{6}=17.93~m$

5 0

3 years ago