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4 years ago

10

6th grade science help me please (:

1 answer:

Alex Ar [27]4 years ago

7 0

Answer:

d

Explanation:

evaporation is the process by which liquid mollecules are converted to gaseous mollecules

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50.0 grams of water vapor at STP occupies how much volume?

oksian1 [2.3K]

Answer:

22.4 L

hope it will help you

6 0

3 years ago

PLEASE HELP ASAPPPP

Liono4ka [1.6K]

So what you do is you write out each element separately and how many of that element on either side of the arrow S-8|S-1
O-2|O-2
on one side theres 8 (S) and 2 (O) and 1(S) and 2(O) if you put a number in front to balance it you are multiplying it the elements number like
SO2 if I put a 2 in front now I will have 2(S) and 4(O)

Answer is 1,8,8

8 0

4 years ago

What volume of carbon dioxide is produced when 0.489 mol of calcium carbonate reacts completely according to the following react

Vilka [71]

Answer:

11.0L of carbon dioxide is produced

Explanation:

Balanced equation: $CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$

According to balanced equation, 1 mol of $CaCO_{3}$ produces 1 mol of $CO_{2}$

So, 0.489 mol of $CaCO_{3}$ produces 0.489 mol of $CO_{2}$

Let's assume $CO_{2}$ behaves ideally.

So, $P_{CO_{2}}V_{CO_{2}}=n_{CO_{2}}RT$

where P is pressure, V is volume , n is number of moles, R is gas constant and T is temperature in kelvin

Plug-in all the values in the above equation-

$(1atm)\times V_{CO_{2}}=(0.489mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

or, $V_{CO_{2}}=11.0L$

So, 11.0L of carbon dioxide is produced

4 0

3 years ago

The average molecular weight for element X is 59.97 g/mol. There are two known isotopes of element X, one weighing 59 g/mol, and

Temka [501]

Answer:

The correct answer is:48.5% X-59, 51.5% X-61

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)

Let the fractional abundance of X-59 isotope be 'x'. So, fractional abundance of X-61 isotope will be '1 - x'

For X-59 isotope:

Mass of X-59 isotope =59 g/mol

Fractional abundance of X-59 isotope = x

For X-61 isotope:

Mass of X-61 isotope = 61 g/mol

Fractional abundance of X-61 isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$59.97 g/mol=[(59 g/mol\times x)+(61 g/mol\times (1-x))]\\\\x=0.515$

Percentage abundance of X-59 isotope = $0.515\times 100=51.5\%$

Percentage abundance of X-61 isotope = $(1-0.515)=0.485\times 100=48.5\%$

Hence, the percentage abundance of both the isotopes X-59and X-61 are 51.5% and 48.5% respectively.

4 0

3 years ago

How can water be decomposed into its elements, hydrogen and oxygen

algol [13]

Water can be decomposed with electrolysis. Which gives you oxygen and hydrogen with their separate types of atoms.

4 0

4 years ago