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2 years ago

What would the products of a double-replacement reaction between Na2S

Physics

1 answer:

MAVERICK [17]2 years ago

8 0

Answer:

A. NaF and MgS

Explanation:

The reaction equation is given as:

Na₂S + MgF₂ → 2NaF + MgS

The product of this reaction is NaF and MgS.

In a double displacement reaction there is an actual exchange of partners to form new compounds. This reaction takes place between ionic compounds.

The driving force for the reaction is:

formation of an insoluble precipitate
formation of water or any other non-ionizing compound
liberation of a gaseous product.

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The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple

lutik1710 [3]

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

Explanation:

4 0

2 years ago

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a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

shutvik [7]

Isothermal Work = PVln(v₂/v₁)

PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = $e^{0.3098}$

v₂ = 19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L Option b.

6 0

2 years ago

g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the

ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m

sin θ = m λ / a

sin θ = 3 630 10⁻⁹ / 5 10⁻⁶

sin θ = 3.78 10⁻¹ = 0.378

θ = sin⁻¹ 0.378

to better see the result let's find the angle in radians

θ = 0.3876 rad

let's reduce to degrees

θ = 0.3876 rad (180º /π rad)

θ = 22.2º

4 0

2 years ago

PLS HELP DUE AT 10 PM

Rina8888 [55]

I believe it is C hope i helped!

6 0

3 years ago

Read 2 more answers

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

6 0

3 years ago