All categories

4 years ago

What is the mass of 600.00 mL sample of seawater with a density of 1.025 g/mL?

Chemistry

1 answer:

kvasek [131]4 years ago

5 0

Answer:

<h2>The answer is 615 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of seawater = 600 mL

density = 1.025 g/mL

The mass is

mass = 600 × 1.025

We have the final answer as

Hope this helps you

You might be interested in

Which of the following can be characterized as a physical change?

Taya2010 [7]

Answer:

D. Sweat evaporates from your skin

Explanation:

A physical change is a reversible process, no new substance is formed. A physical change does not affect the chemical composition of a substance.

Sweat evaporation from the skin is a physical change, because the change that occur is merely a change of state, Liquid to Gaseous state, no new substance is formed.

7 0

4 years ago

How many moles of silicon (Si) are in 1.50 x 1023 atoms Si?

erik [133]

Answer:

1534.5

Explanation:

you just multiply i think

8 0

3 years ago

Help on this question

Rufina [12.5K]

Answer:

Explanation:

Hopefully this helps.

4 0

3 years ago

How many unchanged atoms would remain after 3 half-lives if the initial sample had 600 atoms? Explain.

Leviafan [203]

75 divide 600/2 300/2 150/2

7 0

4 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

<u>Answer:</u> The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

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)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago