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Elodia [21]
3 years ago
9

The mass of an object is 18g and its volume is 9 cm3 Sink Float​

Chemistry
1 answer:
11111nata11111 [884]3 years ago
6 0
The answer is:
It will float
You might be interested in
How many particles of Na are there in 1.43g of a molecular compound with a Molar mass of 23g?
Olin [163]

Answer:

3.74 x 10²² particles

Explanation:

Given parameters:

Mass of compound  = 1.43g

Molar mass of compound  = 23g

Unknown:

Number of particles of sodium = ?

Solution:

To find the number of particles of Na in the compound, we need to obtain the mass of sodium from the total mass given;

          Mass of sodium  = \frac{molar mass of Na}{molar mass of compound} x mass of sample

                                      = \frac{23}{23}  x 1.43g

                                       = 1.43g

Now find the number of moles of this amount of Na in the sample;

          Number of moles  = \frac{mass}{molar mass} = \frac{1.43}{23}  = 0.062mole

Now;

                    1 mole of substance  = 6.02 x 10²³ particles

                       0.062 mole of substance  =  0.062 x 6.02 x 10²³ particles

                                                                     = 3.74 x 10²² particles

7 0
3 years ago
He chemical formula for lithium hydride is LiH A chemist determined by measurements that 0.0800 moles of lithium hydride partici
sweet-ann [11.9K]

Answer:

Mark me brainlist I will then answer in comment

3 0
3 years ago
Soluble ionic compounds that dissociate naerly completely when dissolved are classified as _______.
Anettt [7]

The answer is D. Strong electrolytes

Strong electrolyte is a solute or solution that completely or almost  completely dissociates when in solution. These are good conductors of electricity only  when in molten/aqueous solution.

Strong electrolyte(aq) → Cation+(aq) + Anion−(aq)


6 0
3 years ago
A mixture of caco3 and (nh4)2co3 is 60.7 % co3 by mass. part a find the mass percent of caco3 in the mixture.
blagie [28]

The mixture contains:

CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass

Let, the total mass = 100 grams

Mass of CaCO3 = x grams

Mass of (NH4)2CO3 = y grams

Thus,       x + y = 100 ------------(1)

Mass of CO3 = 60.7% = 60.7 g

Molar mass of CO3 = 60 g/mol

Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles

The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,

moles CaCO3 + moles (NH4)2CO3 = 1.012

mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012

x/100 + y/96 = 1.012---------(2)

based on equation 1 we can write: y = 100-x

x/100 + (100-x)/96 = 1.012

x = 71.2 g

Mass of CaCO3 = 71.2 g



7 0
4 years ago
A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b
sveta [45]

Answer:

  • Part A: m = 0.02356 mol/kg = 0.02356 m
  • Part B: Xsolute = 4.243×10⁻⁴
  • Part C: % m/m = 0.1376%
  • Part D: ppm = 1,376 ppm

Explanation:

<u>1. Data:</u>

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

<u>2. Formulae</u>

  • M = n solute / V sol (L)
  • m = n solute / Kg solvent
  • X solute = n solute / N total
  • % m/m = (mass of solute / mass of solution) × 100
  • ppm = (mass of solute / mass of solution) × 1,000,000
  • density = mass in grams / volume in mL

<u>3. Solution</u>

<u>Part A: Calculate the molality of the salt solution. </u>

<u />

            m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg  H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

<u>Part B: Calculate the mole fraction of salt in this solution</u>.

          X solute = n solute / N total

i) n solute =  2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

<u>Part C: Calculate the concentration of the salt solution in percent by mass</u>.

         % m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

       ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0
3 years ago
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