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3 years ago

Calcium has a charge of +2. The chart lists the charges of different ions.

Physics

2 answers:

vazorg [7]3 years ago

8 0

Answer: $CaO$ , $CaF_2$ , or $CaCl_2$

Explanation:- For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here calcium metal is having an oxidation state of +2, thus the anions must bear a negative charge to form neutral ionic compound.

Thus the possibilities are:

$Ca^{2+}+O^{2-}\rightarrow CaO$

$Ca^{2+}+P^{3-}\rightarrow Ca_3P_2$

$Ca^{2+}+Cl^{-}\rightarrow CaCl_2$

$Ca^{2+}+F^{-}\rightarrow CaF_2$

dexar [7]3 years ago

3 0

The possible resulting chemical formulas for an ionic compound with calcium given the respective charges of the ions are: CaO, CaMg, or CaF₂ and CaO, CaF₂, or CaCl₂. This is because when dealing with these compounds, you simply need to interchange the oxidation state of the two elements and place as the subscript of the element. For instance, when we have Ca²⁺ and F⁻, the result is CaF₂. However, when the oxidation states of the two compounds are equal, the subscript is 1. That is, for Ca²⁺ and Mg²⁻, the result is CaMg. And for Ca²⁺ and Cl⁻, the result is CaCl₂.

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A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V

yan [13]

<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

Across 10.0Ω = 2.6Ω

Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃ -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

<em>Substitute these values into equation (i) as follows;</em>

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = $\frac{9.0}{35.0}$

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1.<em> 5.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2.<em> 10.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3.<em> 20.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0

3 years ago

Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su

kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

$F_g = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

7 0

2 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =\frac{d}{t} (3)$