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2 years ago

Which is a correct method used in a scientific experiment involving acid and base solutions

Physics

1 answer:

Nuetrik [128]2 years ago

3 0

<span> Adding an indicator to a solution to test its acidity. There weren't options so hope it helps.</span>

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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature $T_i$ = 93.2°C

We know

$Q = m c \Delta T$ ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in $J/g^{\circ}C$

$\Delta T$ = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = $125\times 0.9\times (93.2-T_f)$

Heat gained by the water = $100\times 4.184\times(T_f -18.3)$

thus, we have

$125\times 0.9\times (93.2-T_f)$ = $100\times 4.184\times(T_f -18.3)$

$10485-112.5T_f = 418.4T_f - 7656.72$

⇒ $T_f = 34.17^oC$

Therefore, the final temperature will be = 34.17°C

6 0

2 years ago

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

padilas [110]

Answer:

a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

tan θ = sin θ / cos θ = sin θ

We replace

d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

d = λ L / y

d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

λ = d y / L

λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

λ = 269.19 10⁻⁹ m

4 0

2 years ago

A force of 200 N is applied on small piston of a pascal press. What would be the

VladimirAG [237]

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

$P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}$

Where;

F₂ is the force on big piston

$F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N$

Therefore, the force applied on the big piston is 1306.67 N

3 0

1 year ago

What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u

Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 × $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

$E=h\nu_{o}$ ..................1

here E is the energy of electron per atom $E=\frac{x}{N}$ and h is plank constant i.e. $6.626\times10^{-34} Js$ and x is binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$

E = $3.19\times10^{-19} J$

so put value in equation 1 we get

$\frac{x}{6.023\times10^{23}}$ = 2.50 × $10^{14}$ × $6.626\times10^{-34} Js$

solve it we get

x = 99770.99

so binding energy is 99771 J/mol

4 0

2 years ago

A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it

Eduardwww [97]

Part b is equal to F in standards of society and it’s quality of math during the 1900s
(That was a bit of Social Studies lol)

4 0

2 years ago