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3 years ago

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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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Calculate the pressure of the CO₂ (g) in the container at 425 K.

Elanso [62]

The pressure of the CO₂ = 0.995 atm

<h3>Further explanation</h3>

The complete question

<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>

<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>

<em />

<em />

Gay Lussac's Law

When the volume is not changed, the gas pressure is proportional to its absolute temperature

$\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

P₁=0.7 atm

T₁=299 K

T₂=425 K

$\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{0.7\times 425}{299}=0.995 `atm$

<em />

6 0

3 years ago

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

3 years ago

What does the peak of a probability curve for an electron in an atom indicate?

Eduardwww [97]

The distance from the nucleus at which the electron is most likely to be found

3 0

3 years ago

Read 2 more answers

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

3 years ago

For each of the following sublevels, give the n and l values and the number of orbitals: (a) 6g; (b) 4s; (c) 3d.

olya-2409 [2.1K]

Answer:

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 5

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 2l+1 = 5

4 0

3 years ago