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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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An oblique rectangular prism with a square base has a volume of 539 cubic units. The edges of the prism measure 7 by 7 by 14 uni

Gala2k [10]

Answer:

3 units

Solution:

V=539 cubic units

Square base, with edge a=7 units

Slanted edge length: s=14 units

V=Ab h

Ab=49 square units

539 cubic units = (49 square units) h

h= 11 units

s-h=14 units-11 units

s-h=3 units

8 0

3 years ago

Read 2 more answers

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .

8 0

3 years ago

30 points - how do I do 2 b and c?

Delvig [45]

1. its must be B and 2. must be C

7 0

3 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

3 years ago

The turntable in a microwave oven has a moment of inertia of 0.039 kg⋅m2 and is rotating once every 4.4 s . Part A What is its k

gayaneshka [121]

To solve this problem it is necessary to apply the concepts related to Kinetic Energy, specifically, since it is a body with angular movement, the kinetic rotational energy. Recall that kinetic energy is defined as the work necessary to accelerate a body of a given mass from rest to the indicated speed.

Mathematically it can be expressed as,

$KE = \frac{1}{2} I\omega^2$