Answers
An alloy, a metallic substance composed of two or more elements, as either a compound or a solution is a homogeneous mixture. The correct option is d.
What is an alloy?
An alloy is a mixture of chemical elements of which at least one is a metal. Unlike chemical compounds with metallic bases, an alloy will retain all the properties of metal in the resulting material, such as electrical conductivity, ductility, opacity, and lustre, but may have properties that differ from those of pure metals, such as increased strength or hardness.
In some cases, an alloy may reduce the overall cost of the material while preserving important properties. In other cases, the mixture imparts synergistic properties to the constituent metal elements such as corrosion resistance or mechanical strength. Alloys are defined by a metallic bonding character. The alloy constituents are usually measured by mass percentage for practical applications, and in atomic fraction for basic science studies.
Alloys are usually classified as substitutional or interstitial alloys, depending on the atomic arrangement that forms the alloy.
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Answer:
An alloy is a homogeneous mixture
Explanation:
D
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Answers
Answer:
You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate
Explanation:
First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.
Answers
Answer:
x(t) = −39e
−0.03t + 40.
Explanation:
Let V (t) be the volume of solution (water and
nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid
measured in liters after t minutes, and let c(t) be the concentration (by volume) of
nitric acid in solution after t minutes.
The volume of solution V (t) doesn’t change over time since the inflow and outflow
of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is
c(t) = x(t)
V (t)
=
x(t)
200
.
We model this problem as
dx
dt = I(t) − O(t),
where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,
both measured in liters of nitric acid per minute. The input rate is
I(t) = 6 Lsol.
1 min
·
20 Lnit.
100 Lsol.
=
120 Lnit.
100 min
= 1.2 Lnit./min.
The output rate is
O(t) = (6 Lsol./min)c(t) = 6 Lsol.
1 min
·
x(t) Lnit.
200 Lsol.
=
3x(t) Lnit.
100 min
= 0.03 x(t) Lnit./min.
The equation is then
dx
dt = 1.2 − 0.03x,
or
dx
dt + 0.03x = 1.2, (1)
which is a linear equation. The initial condition condition is found in the following
way:
c(0) = 0.5% = 5 Lnit.
1000 Lsol.
=
x(0) Lnit.
200 Lsol.
.
Thus x(0) = 1.
In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is
µ(t) = exp Z
P(t) dt
= exp
0.03 Z
dt
= e
0.03t
.
The solution is
x(t) = 1
µ(t)
Z
µ(t)Q(t) dt + C
= Ce−0.03t + 1.2e
−0.03t
Z
e
0.03t
dt
= Ce−0.03t +
1.2
0.03
e
−0.03t
e
0.03t
= Ce−0.03t +
1.2
0.03
= Ce−0.03t + 40.
The constant is found using x(t) = 1:
x(0) = Ce−0.03(0) + 40 = C + 40 = 1.
Thus C = −39, and the solution is
x(t) = −39e
−0.03t + 40.
Answers
Answer:
c
Explanation:
Answer:
c
Explanation:
Answers
Answer:
2.14x10¹⁹ atoms of Cu were deposited
Explanation:
The increased in mass of the glass slide is due the deposition of copper.
That means the mass of copper deposited is 2.26mg = 2.26x10⁻³g Cu
To know the copper atoms we need to convert this mass to moles of Cu using molar mass of copper (63.546g/mol), and these moles are converted to atoms using Avogadro's number (6.022x10²³ atoms = 1 mole)
Moles Cu:
2.26x10⁻³g Cu * (1 mol / 63.546g) = 3.556x10⁻⁵ moles Cu
Atoms Cu:
3.556x10⁻⁵ moles Cu * (6.022x10²³ atoms / 1 mole) =
2.14x10¹⁹ atoms of Cu were deposited
O True
O False
Answers
thank me later
Answer:
False
Explanation:
It depends on wixh kid of light is it
Example
Sun light difders from lamp light
B. 100 mL volumetric flask
C. 10 mL volumetric pipet
D. 25 mL volumetric pipet
E. 10 mL graduated pipet
F. 25 mL graduated pipet
Answers
Answer:
C
Explanation:
The volumetric pipet is designed for a specific volume of solution. In this case, a 10 mL of volumetric pipet is required to take exactly 10 mL of the solution 1.00 M and add into a volumetric flask of 100 mL. Then water or solvent(s) can be added to reach the ring of the volumetric flask. This volumetric flask is also a laboratory equipment for precis dilutions beside the volumetric pipet