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vodomira [7]
3 years ago
14

A radioactive isotope has a half-life of 3.0 hours. If a scientist has 30 grams of the isotope, how much is left after 15 hours?

Chemistry
1 answer:
dmitriy555 [2]3 years ago
7 0

Answer:

0.9375

Explanation:

3.0x30x15 = 17 496 000 kg sx^{2}

subtract or divde

then you take the 9 and 375 add 0.9000 and combine them and there you go.

                                                          +\frac{0.9000}{375} =0.9375

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In the metric system, the ________ describes the relationship of each unit to the base unit.
gavmur [86]

Answer:

prefix

Explanation:

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3 years ago
How are the Spectra of the gasses Different?
galina1969 [7]

When gasses are heated they emit only certain wavelengths of light (an emission line spectrum). Different gasses emit different wavelengths. A cool object (gas or solid) can absorb some of the light passing through it. The temperature of an object is a measure of how much energy its atoms have.

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3 0
3 years ago
Calculate the mass (g) of 1.5 moles of Sulfur. ​
just olya [345]

Answer:

48 g S

Explanation:

Step 1: Define

Molar Mass of Sulfur (S) - 32.07 g/mol

Step 2: Use Dimensional Analysis

1.5 \hspace{3} mol \hspace{3} S(\frac{32.07 \hspace{3} g \hspace{3} S}{1 \hspace{3} mol \hspace{3} S} ) = 48.105 g S

Step 3: Simplify

We have 2 sig figs.

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7 0
3 years ago
I just started learning about kinetic molecular theory, and I’m not sure how to answer the question circled below
lions [1.4K]

Answer : The value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}

Solution : Given,

At STP conditions,

Pressure = 1 atm

Temperature = 273 K

Number of moles = 1 mole

Volume = 22.4 L

Formula used :     R=\frac{PV}{nT}

where,

R = Gas constant

P = pressure of gas

T = temperature of gas

V = volume of gas

n = number of moles of gas

Now put all the given values in this formula, we get the values of 'R'.

R=\frac{(1atm)\times (22.4L)}{(1mole)\times (273K)}

R=0.0821\text{ L atm }mol^{-1}K^{-1}

Therefore, the value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}.

7 0
3 years ago
If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?
Anastasy [175]
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law 
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>

<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>

<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol.  </span>

<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>

<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
8 0
3 years ago
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