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enot [183]
4 years ago
12

The half life or radium 226 is 1630 years

Chemistry
1 answer:
ELEN [110]4 years ago
5 0

So what is the question?

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A lead mass is heated and placed in a foam cup caloriemeter containing 40.0mL of water at 17.0 C. The water reaches a temperatur
Mrrafil [7]

the Answer is             502 Joules

8 0
3 years ago
What is the molarity of a sodium chloride solution made by dissolving 4.512 moles to make 2.0 L?
Svet_ta [14]

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

mL

. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

1 L

=

10

3

mL

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

1

, which means that you're going to have to convert the mass of the sample from milligrams to grams

1 g

=

10

3

mg

Sodium chloride,

NaCl

, has a molar mass of

58.44 g mol

−

1

, which means that your sample will contain

unit conversion



280.0

mg

⋅

1

g

10

3

mg

⋅

molar mass



1 mole NaCl

58.44

g

=

0.004791 moles NaCl

This means that the molarity of the solution will be

c

=

n

solute

V

solution

c

=

0.004791 moles

2.00

⋅

10

−

3

L

=

2.40 M

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.

6 0
3 years ago
A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work
DENIUS [597]
Data:
V_{initial} = 590\:mL
T_{initial} = -55.0^0C
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → T_{initial} = 218.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 30.0^0C
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → T_{final} = 303.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }
Product of extremes equals product of means:
218.0* V_{f} = 590*303.0
218.0 V_{f} = 178770
V_{f} = \frac{178770}{218.0}
\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark
7 0
3 years ago
You are performing an experiment that uses 114Ag. 113Ag is radioactive, decays by beta-- emission and has a half-life of 21 minu
maxonik [38]

Answer: The 234.74 grams of sample should be ordered.

Explanation:

Let the gram of  114 Ag to ordered be N_o

The amount required for the beginning of experiment = 0.0575 g

Time requires to ship the sample = 4.2hour = 252 min(1 hr = 60 min)

Half life of the sample ={t_{\frac{1}{2}} = 21 min

\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{21 min}=0.033 min^{-1}

\log[N]=\log[N_o]-\frac{\lambda t}{2.303}

\log[0.0575 g]=\log[N_o]-\frac{0.033 min^{-1}\times 252 min}{2.303}

N_o=234.74 grams

The 234.74 grams of sample should be ordered.

4 0
3 years ago
To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7
Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0
3 years ago
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