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3 years ago

An example of a radioactive isotope is waht

1 answer:

8 0

Like carbon-12 (C-12) and carbon-14 (C-14). C-12 is not radioactive but C-14 is radioactive. C-14 is radioactive isotope of C-12.

You might be interested in

What products would you expect from the reaction of ammonia and sulfuric acid in aqueous solution? 2nh3(aq)+h2so4(aq)→ ?

arsen [322]

From the reaction of ammonia and sulfuric acid in aqueous solution 2nh3(aq)+h2so4(aq)→ (NH4)2SO4 + H2O will be formed.Sulfuric acid is diprotic so is able to give up 2 H+ ions. It is an acid-base neutralisation reaction forming ammonium sulphate as the salt. 2NH3 with H2SO4 reacts in a neutralization reaction to form salt water, with ammonium sulphate left behind to crystallize after evaporation.

5 0

3 years ago

A mechanic needs a radiator to have 50% antifreeze solution. the radiator currently is filled with 4 gallons of 15% antifreeze s

Zanzabum

let x=gallons of current mixture to be drained and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.

0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal

Thus, 1.65 gallons of current mixture to be drained and replaced with pure antifreeze.

5 0

3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

68.3g

Explanation:

1. Word equation:

mercury(II) oxide → mercury + oxygen

2. Balanced molecular equation:

2HgO → 2Hg + O₂(g)

3. Mole ratio

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

4. Calculate the number of moles of O₂(g)

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

5. Calculate the number of moles of HgO

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

6. Convert to mass

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

3 years ago

what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c

slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

$Q = m \times C \times \bigtriangleup T$

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

$C = \frac{Q}{m \times \bigtriangleup T}$

Let us plug in the values in above equation.

$C = \frac{300J}{267g \times 12 C}$

$C = \frac{300J}{3204 g C}$

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0

3 years ago