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iris [78.8K]
3 years ago
10

Chemistrys forerunners​

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
4 0

Answer:

huhhhhhhhhhhhhhhhhhhh??

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each element is designated by its __________ which is usually from the first letters of the elements name
klemol [59]

Answer:

each element is designated by its <u>Chemical Symbol </u>which is usually from the first letters of the elements name

Explanation:

7 0
3 years ago
An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

7 0
3 years ago
Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th
satela [25.4K]

Answer:

General equation for a double-displacement reaction:  

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄  

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄  

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄  

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃  

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃  

NaOH + AgNO₃   --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃  --> NaNO₃ +  Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0
3 years ago
Fatty acids spread spontaneously on water to form a monomolecular film. A solution containing 0.10 mm^3 of a fatty acid is dropp
lara31 [8.8K]

Divide the volume by the area. Using scientific makes things a bit cleaner.

0.10\,\mathrm{mm}^3 = 10^{-1}\,\mathrm{mm}^3

400.\,\mathrm{cm}^2 = 4\times10^2\,\mathrm{cm}^2

Then

\dfrac{10^{-1} \,\mathrm{mm}^3}{4\times10^2\,\mathrm{cm}^2} \cdot \dfrac{\left(\frac{1\,\rm m}{10^3\,\rm mm}\right)^3}{\left(\frac{1\,\rm m}{10^2\,\rm cm}\right)^2} = \dfrac{10^{-1}\times10^{-9} \,\mathrm m^3}{4\times10^2\times10^{-4}\,\mathrm m^2} = \dfrac{10^{-10}}{4\times10^{-2}}\,\mathrm m \\\\ ~~~~~~~~= 0.25\times10^{-8}\,\mathrm m

Now, 1 m = 10⁹ nm, so

0.25 \times10^{-8}\,\mathrm m \cdot \dfrac{10^9\,\rm nm}{1\,\rm m} = 0.25\times10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}

8 0
1 year ago
How many grams of C3H8 would be needed to produce 6.39 grams of CO2?
just olya [345]

Answer:

2.13g

Explanation:

Atomic mass of CO2 =  12 + 32 = 44g/Mol

Atomic mass of C3H8 = 36 + 8 = 44g/Mol

Reaction

C3H8 + 5O2 --> 3CO2 + 4H2O

3CO2 = 6.39g

Required C3H8 = (6.39/(44 x 3)) x 44 = 2.13g  

8 0
3 years ago
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