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ZanzabumX [31]
2 years ago
6

Berny has a 0.15-kilogram candle, a 1.00-kilogram jar, and a 0.25 kilogram candle holder. He places the candle and candle holder

in the jar, lights the candle, and then immediately closes the jar.
What should the total mass of the jar and its contents be after the candle goes out? A. 1.40 kilograms B. 1.25 kilograms C. 1.00 kilograms D. 1.15 kilograms
Chemistry
2 answers:
Kobotan [32]2 years ago
5 0
It's a because if you add them together you till get 1.40
Lera25 [3.4K]2 years ago
4 0

ITS A :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)

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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
1 year ago
Determine the number of moles and mass requested for each reaction in Exercise 4.42.
suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

  1. Step 1: Write the balanced equation
  2. Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
  3. Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.

Step 3:

10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}

10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.

Step 3:

1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}

1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.

Step 3:

128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}

128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.

Step 3:

20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2

[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2[/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.

Step 3:

1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}

5 0
1 year ago
How many Ca2+ ions are in a sample of CaSO4 having a total mass of 68.07 g?
harkovskaia [24]

Answer:

There are  

17.01

Explanation:

The chemical formula for calcium phosphate is  

Ca

3

(PO

4

)

2

. This means that in one mole of calcium phosphate, there are three calcium ions and two phosphate ions.

6 0
1 year ago
Rust results from iron’s reaction to oxygen. An iron nail gains mass when it rusts. How does this reaction support the law of co
Olegator [25]

Answer:

Option (A). The mass of the rusted nail equals the mass of iron and the oxygen from the air it reacted with to form the rust.

5 0
1 year ago
Which statement best describes how electrons fill orbitals in the periodic table? Electrons fill orbitals in order of their incr
Stolb23 [73]
Electrons fill orbitals in order of increasing energy from left to right. As the group number increase also the number of valence electorns of each group will increases
5 0
2 years ago
Read 2 more answers
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