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dedylja [7]
3 years ago
12

Srihari has a torch with him and 2 new batteries. However, the torch requires 3 batteries to work. Is there any way Srihari can

get the torch to glow?
Chemistry
1 answer:
zloy xaker [14]3 years ago
6 0

Answer:

go to the store to buy more batteries

Explanation:

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In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
artcher [175]
Stoichiomety:

1 moles of C + 1 mol of O2 = 1 mol of CO2

multiply each # of moles times the atomic molar mass of the compund to find the relation is weights

Atomic or molar weights:

C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol

Stoichiometry:

12 g of C react with 32 g of O2 to produce 44 g of CO2

Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen

And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.

You cannot obtain 72 g of CO2 from 18 g of C.

May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
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In the water cycle, matter moves against the pull of gravity during
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Match each statements with the state of matter it describes?
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Solid  liquids , plasma gas

Explanation:

It retains its shape regardless of the shape of the container

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Which of the following is an example of CHEMICAL weathering? (select all
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You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol
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Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH

Step 4: Calculate the molarity of the acetic acid solution

M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g

Step 6: Calculate the percentage of acetic acid in the solution

\frac{0.4186g}{25.00mL}  \times 100\% = 1.674 \%(m/V)

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