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2 years ago

A silver necklace has a density of 10.5 g/mL and a volume of 1.5 mL. What is the mass of the silver necklace?

Chemistry

1 answer:

melamori03 [73]2 years ago

4 0

Answer:

15.75g

Explanation:

Rearrange the formula: density = mass/volume, so that mass is the subject: mass = density × volume

Now substitute (plug in) the known values into the formula:

mass = 10.5 × 1.5 ⇒ mass = 15.75g

The mass of the silver necklace is 15.75g

Hope this helps!

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In 3–5 sentences, explain the various factors that should be considered when implementing green roofs

Kruka [31]

Answer:

you should consider waterproofing. depending on the climate in your area, you need to apply multiple layers of waterproofing in the roof for it to support vegetation. you should also confused the types of plants. the types of plants is important for the success of your roof. if you live in a dry area, you don't want to choose plants that can live in dry areas without drying out.

6 0

3 years ago

Create a graphic representation that illustrates the steps from gene to protein. Label and describe each step.

Vesna [10]

Okay so,

1) Translation- show the RNA strand attatching to a DNA strand with the complimentary base pairs. introns are spliced

2) mRNA leaves the cell and joins with a ribosome

3) Transcription - tRNA (clover shaped) reads each codon (triplets) which each code for an amino acid. The stop codons on the end tell the tRNA that the chain is finished

4) the sequence forms the primary structure (all peptide bonds) which determines the shape of the secondary (hyrdogen and peptide) and hence determines the shape of the tertiary structure of a protein (ionic, hydrogen, disulfide bridges and hydrophibic interactions)

Hope this helps :)

6 0

3 years ago

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda

bezimeni [28]

Answer: The average atomic mass of X is 28.09 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

For isotope 2:

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

For isotope 3:

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]$

$\text{Average atomic mass of X}=28.09amu$

Hence, the average atomic mass of X is 28.09 amu

4 0

3 years ago

A train travels 74 kilometers in 3 hours, and then 73 kilometers in 2 hours. What is its average speed? km/hr

nikitadnepr [17]

12km/Hr is the correct answer

5 0

3 years ago

Read 2 more answers

Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

3 years ago