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Nataly_w [17]
3 years ago
14

A student determined that a 50mL sample of an unknown liquid had a density of 1 g/mL. A 100mL sample of the same liquid would ha

ve a density of 2 g/mL. True or false
Chemistry
1 answer:
Lyrx [107]3 years ago
4 0

Answer: False

Explanation: The density of the liquid will always be the same, no matter how much of the liquid you have.

For example, the density of the water in a pool is the same density as the water in a small cup.

You might be interested in
Change the given word equation into balanced equation <br><br>Hydrogen+Chloride=Hydrogen chloride​
nirvana33 [79]

Answer:

h2 +2cl -> 2hcl

Explanation:

h2 + cl -> HCl

h2 +2cl -> 2hcl

3 0
4 years ago
A piece of copper with a mass of 218 g has a heat capacity of 83.9 j/°c. what is the specific heat of copper?
kari74 [83]
The answer to this question would be: 0.385  j/g°c

Specific heat is the amount of energy needed to increase 1 degree of celcius per one gram of a specific mass. In this question, it needs 83.9 j/°c energy to increase the temperature of 218 g copper. Then, the specific heat would be:  (83.9 j/°c) / 218g= 0.38486  j/g°c
4 0
4 years ago
The enthalpy of combustion of n-octane, C8H18, is –4.79 × 107 J/kg. What is the enthalpy of combustion expressed in kJ/g?
SIZIF [17.4K]

Answer:

47,9 kJ/g

Explanation:

The enthalpy of combustion is the change in energy due the reaction of combustion of n-octane. If the enthalpy is -4,79x10⁷ J/kg And knowing (1000J = 1kJ) (1000g = 1kg). The enthalpy of combustion expressed in kJ/g is:

-4,79x10^7J/kg \frac{1kJ}{1000J} \frac{1kg}{1000g} = <em>47,9kJ/g</em>

<em></em>

I hope it helps!

6 0
4 years ago
what quantity of ethylen glycol must be added to 125 g of water to raise the boiling point of 1.0 degrees celsuis
labwork [276]

Answer:

We should add 15.15 grams of ethylen glycol.

Explanation:

Step 1: Data given

Mass of water = 125 grams

Temperature change = 1.0 °C

The boiling point elevation constant, Kbp, for water is 0.5121 °C/m

Step 2: Calculate mass needed

ΔT = i* Kb *  m

⇒ with i = the van't Hoff factor = 1

⇒ with Kb = 0.5121 °c/ kg/mol

⇒ with m = molality = moles / mass

ΔT = i* Kb *  m  ⇒ 1°C  = 0.5121 °C / kg/mol *   (X/0.125kg)

X = 0.2441 moles

Step 3: Calculate mass of ethylen glycol

Mass ethylen glycol = moles * molar mass

Mass ethylen glycol = 0.2441 moles * 62.068 g/mol

Mass of ethylen glycol = 15.15 grams

We should add 15.15 grams of ethylen glycol.

4 0
3 years ago
Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl is allowed to react with 17.2 g of O2
alina1380 [7]

Answer:

The limiting reactant for this reaction is the HCl

Explanation:

This is my reaction:

4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)

Molar mass O2 = 32 g/mol

Molar mass HCl = 36,45 g/mol

Mass / Molar mass = Moles

Moles HCl : 63,1 g / 36,45 g/m = 1,73 moles

Moles O2: 17,2 g / 32g/m =  0,54 moles

4 moles of HCl react with 1 mol of O2, according to reaction, so

1,73 moles of HCl, are going to react with, how many moles of O2.

4 moles HCl ___ 1 mol O2

1,73 moles HCl ___ (1,73 . 1)/ 4 = 0,43 moles of O2

O2 is my excess reagent because I need 0,43 moles and I have 0,54 so I have moles in excess.

1 mol of O2 are going to react with 4 moles of HCl

0,54 moles of O2 are going to react with, how many moles of HCl ?

1 mol O2 ____ 4 moles HCl

0,54 mol O2 ___ (0,54 . 4)/ 1 = 2,16 moles

I need 2,16 moles to consume my moles of O2, but I only have 1,73 moles, tha's why the HCl is mi limiting reactant.

3 0
3 years ago
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