What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

Answers

Answer 1

Answer:

The volume of 45.6g of silver if the density of silver is 10.5g/mL is 4.342 ml. The correct option is A.

What is Volume?

Volume is the space occupied by a three-dimensional object. The volume of any object can be calculated by dividing the mass by its density. It is a scalar quantity. It is the total weight is that object.

Silver is an element in the periodic table. It is non-metal, and it is used in making ornaments and in medicines. The volume of the solver is calculated, and the mass and density are given.

$\rm{Volume = (mass)/(density)}$

The mass of silver is given, 45.6g

The density of the element is 10.5g/mL

Putting the value in the equation

The density and the mass would be divided.

Volume = 45.6g / 10.5 = 4.342

Thus, the volume of silver is 4.342 ml. The correct option is A.

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Answer 2

Answer:

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

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Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?

Answers

Answer:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^(+2)(aq)+2(PO_4)^(-3)(aq)$

$Ksp=[Sr^(+2)]^3[(PO_4)^(-3)]^2$

Explanation:

Hello,

In this case, for strontium phosphate, we find an ionic equation for its dissociation as shown below:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^(+2)(aq)+2(PO_4)^(-3)(aq)$

Next, the solubility product is found by applying the law of mass action, considering that the solid salt is not considered but just the aqueous species due to heterogeneous equilibrium:

$Ksp=[Sr^(+2)]^3[(PO_4)^(-3)]^2$

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Which electron configuration matches this model?1s22s22p2
1s22s22p3
1s22s22p4
1s22s22p5

Answers

The electronic configuration that matches that of the model is 1s²2s²2p³; option B.

What is electronic configuration of atoms of elements?

Electronic configuration of atoms of elements is the arrangement of electrons in the electron shells or orbitals in atom.

The electronic configuration of an atom depends on the atomic number of the element.

The atomic number of nitrogen is 7. This means that it has seven electrons in its orbitals.

The electronic configuration that matches that of the nitrogen atom is 1s²2s²2p³.

In conclusion, the electronic configuration of nitrogen indicates that it has seven electrons in its orbitals.

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Note that the complete question is given below:

Consider the model of the nitrogen atom. Which electron configuration matches this model?

1s22s22p2

1s22s22p3

1s22s22p4

1s22s22p5

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Answers

Ba(IO₃)₂(s) partially dissociates in water as Ba²⁺(aq) and IO₃⁻(aq).
Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial
Change -X +X 2X
Equilibrium X 2X

Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰ = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

Which phase changes will result in an increase in energy of the substance

Answers

Answer:

when gas condenses to liquid the quantity of energy converts.

Explanation:

two phase changes where the heat in energy is released

potential energy. During a phase change, the heat added (PE increases) or released (PE decreases) will allow the molecules to move apart or come together. Heat absorbed causes the molecules to move farther apart by overcoming the intermolecular forces of

The following reaction was monitored as a function of time: A→B+C A plot of ln[A] versus time yields a straight line with slope −4.3×10−3 /s. If the initial concentration of A is 0.260 M, what is the concentration after 225 s?

Answers

The concentration after 225 s is 0.099 M.

As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.

So, Slope = k = $4.3*10^(-3)/s$

Rate law for first order kinetics:

$t=(2.303)/(k) log (a)/(a-x)$

where,

k = rate constant = $4.3*10^(-3)/s$

t = time passed by the sample = 225 s

a = initial amount of the reactant = 0.260 M

a - x = amount left after decay process = ?

On substituting the values:

$t=(2.303)/(k) log (a)/(a-x)\n\nt=(2.303)/(4.3*10^(-3)) log (0.260)/(a-x)\n\na-x=0.099M$

Therefore, the concentration after 225 s is 0.099 M.

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Answer : The concentration after 225 s is, 0.099 M

Explanation :

As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.

So, Slope = k = $4.3* 10^(-3)s^(-1)$

Expression for rate law for first order kinetics is given by:

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant = $4.3* 10^(-3)s^(-1)$

t = time passed by the sample = 225 s

a = initial amount of the reactant = 0.260 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$225=(2.303)/(4.3* 10^(-3))\log(0.260)/(a-x)$

$a-x=0.099M$

Therefore, the concentration after 225 s is, 0.099 M

Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ

Answers

Explanation:

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