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Fynjy0 [20]
3 years ago
10

Which milligram quantity contains a total of four significant figures

Chemistry
2 answers:
denis23 [38]3 years ago
7 0
D has a total of four significant figures.
Ksenya-84 [330]3 years ago
3 0

Answer:

D

Explanation:

The first 0 is sandwiched between the 3&1 and a ending 0 that is after the decimal is significant.

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Theyre temporary magnets, so they can be turned on and off.
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3 years ago
Read 2 more answers
How many electrons in an atom can have each of the following quantum number or sublevel designations?
lutik1710 [3]

Answer: a:6 electrons, b:10electrons, c: 2electrons

Explanation:

1.Principle quantum number(n)

n is the number of shell, it is a positive integer. The maximum number of electrons in a shell is 2(n2)

2.Azimuthal quantum number(l)

l is the number of subshell in the principal shell

The name of subshell are s, p, d, f

The number of nodes in each subshell is

s is l=0, p is l=1, d is l=2, f is l=3.

Each shell can have 2 x l + 1 sublevels, and each sublevel can accommodate maximum of 2 electrons and have two electrons.

A. n=2, l=1 then 2 x 1 + 1= 3 subshell, 3*2= 6 electrons in the p subshell

b. 3d, d is l = 2 then 2 x 2 + 1 = 5 sublevels, 5*2=10 electrons in the d subshell

c. 4s, s is l=0 then 2 x 0 + 1 = 1 sublevel, 1*2=2 electrons in the s subshell.

6 0
3 years ago
Name:
Liono4ka [1.6K]

Answer:

1. 505g is the mass of the aluminium.

2. The answer is in the explanation

Explanation:

1. To solve this question we need to find the volume of the rectangle. With the volume and density we can find the mass of the solid:

Volume = 7.45cm*4.78cm*5.25cm

Volume = 187cm³

Mass:

187cm³ * (2.702g/cm³) = 505g is the mass of the aluminium

2. When the temperature of a liquid increases, the volume increases doing the density decreases because density is inversely proportional to volume. And works in the same way for gases because the temperature produce more collisions and the increasing in volume.

4 0
3 years ago
Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the
expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

5 0
3 years ago
Which level of protein structure do beta-pleated sheets and alpha helices represent?
pychu [463]
The primary structure is the amino acids' unique sequence. The polypeptide's local folding to form structures such as the α-helix and β-pleated sheet constitutes the secondary structure. The overall three-dimensional structure is the tertiary structure
3 0
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