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ELEN [110]
3 years ago
6

A silver necklace is made from a pure sample of silver that is free of I'm purities. Which term or terms could be used to descri

be this sample of silver ?
Chemistry
1 answer:
geniusboy [140]3 years ago
5 0
The terms that apply to this silver necklace are:
pure chemical substance
element

The necklace consists purely of silver, making it a pure chemical substance; moreover, silver itself is an element, not a compound. Therefore, the term element is also fitting for the necklace.
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One gallon of water has a mass of 3.78kg. Water is 11.2% hydrogen by mass. Calculate the mass of hydrogen contained in this gall
olchik [2.2K]

Answer:

A model is developed for predicting oxygen uptake, muscle blood flow, and blood chemistry changes under exercise conditions. In this model, the working muscle mass system is analyzed. The conservation of matter principle is applied to the oxygen in a unit mass of working muscle under transient exercise conditions. This principle is used to relate the inflow of oxygen carried with the blood to the outflow carried with blood, the rate of change of oxygen stored in the muscle myoglobin, and the uptake by the muscle. Standard blood chemistry relations are incorporated to evaluate venous levels of oxygen, pH, and carbon dioxide.

Explanation:

8 0
3 years ago
Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?
Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06

6 0
3 years ago
Matching
kotegsom [21]

Answer:

a mixture of molecules - Box f

atoms of a pure elementa metal - Box D

a solid compound - Box C

a mixture of elements - Box A

Explanation:

Box a has mixture of elements which forms a solid like shape but there are different elements present in the box. The box f has mixture of molecules in which many atoms are combined together. Box c has solid compound with single element.

5 0
3 years ago
The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri
levacccp [35]

<u>Answer:</u> The equilibrium concentration of HCl is 2.26\times 10^{-3}M

<u>Explanation:</u>

We are given:

Moles of NH_4Cl(s) = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Molarity of NH_4Cl=\frac{0.564}{1}=0.564M

The given chemical equation follows:

                  NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)

<u>Initial:</u>         0.564

<u>At eqllm:</u>     0.564-x          x              x

The expression of K_c for above equation follows:

K_c=[NH_3][HCl]

The concentration of pure solid and pure liquid is taken as 1.

We are given:

K_c=5.10\times 10^{-6}

Putting values in above equation, we get:

5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M

Negative sign is neglected because concentration cannot be negative.

So, [HCl]=2.26\times 10^{-3}M

Hence, the equilibrium concentration of HCl is 2.26\times 10^{-3}M

5 0
3 years ago
What is the mass of 3.20x10^23 formula units of iron (III) oxide (Fe2O3)?
yaroslaw [1]

The mass of iron (III) oxide (Fe2O3) : 85.12 g

<h3>Further explanation</h3>

Given

3.20x10²³ formula units

Required

The mass

Solution

1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol of Fe₂O₃ :

\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532

mass of Fe₂O₃ (MW=160 g/mol)

\tt mass=mol\times MW=0.532\times 160=85.12~g

4 0
2 years ago
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