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3 years ago

5

7.7. A painter leans his back against a painted wall while looking into a 1m long mirror at the opposite end of a rectangular ro

om as shown in the given figure. How much of the painted wall can he see through the given mirror?
a. 1m
b. 2m
c. 12m
d. 6m

1 answer:

Karolina [17]3 years ago

8 0

Answer:

Correct answer is b) 2m

Explanation:

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How does an increase in temperature affect a chemical reaction??

blsea [12.9K]

An increase in the temperature will speed up the reaction by increasing the frequency and efficiency of the collisions of molecules.

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what is the percent yield of NaCl if 31.0 g of CuCl reacts with excess NaNo3 to produce 21.2 g of NaCl

olchik [2.2K]

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

Balanced Equation: CuCl + NaNO₃   → NaCl + CuNO₃

Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol

the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,

then moles of NaCl = 0.31 mol

Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g

⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.

Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %

NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.

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You have 400 kg of a radioactive substance with a short half-life of 1000 years. how much will be left after these times?

Amiraneli [1.4K]

Half-life of a radioactive substance is the time required to reduce the amount of substance to half of its initial amount.

In present case, half-life is material is given as 1000 years and initial amount of material is given as 400 kg

Answer 1) Since, half-life of radio-active substance is 1000 years, therefore after 1st half life, amount of the material will be left to half the initial amount. Hence, amount of substance left after 1000 years = 400/2 = 200 kg.

Answer 2) For 2000 years, radioactive material has crossed 2 times the half life. Therefore , amount of the material will be left to 1/4 the initial amount. Hence, amount of substance left after 2000 years = 400/4 = 100 kg.

Answer 3) For 4000 years, radioactive material has crossed 4 times the half life. Therefore , amount of the material will be left to 1/16 the initial amount. Hence, amount of substance left after 4000 years = 400/16 = 25 kg.

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3 years ago

HELP please! Which of these describes how heat is transferred by conduction

VLD [36.1K]

D. Air molecules touch the warm ground, heating them up

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3 years ago

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

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3 years ago