Why would a doctor most likely restrict a patient's contact with other people while the patient receives internal radiation?

Select all of the following which are characteristics of metals.

shtirl [24]

Answer: Metals are malleable. Most metallic elements are solids at room temperature.

Explanation:

8 0

2 years ago

Which of the following decomposition reactions is/are correct?

Valentin [98]

The generalized reaction for chemical decomposition is: AB → A + B

NaOH is sodium hydroxide. When sodium and water is combined it makes sodium hydroxide and hydrogen

When sodium hydroxide decomposes under thermal decomposition, it breaks down into sodium oxide and water.

Thus, C) 2NaOH Na2O + H2O

4 0

2 years ago

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$

Therefore, the leftover of sodium nitrate is:

$m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3$

Finally, the percent yield is computed via:

$Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%$

Best regards!

6 0

2 years ago

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0

3 years ago

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with

Sphinxa [80]

Answer : The work done is, $1.98\times 10^4J$

Explanation :

The given balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

When 4 moles of $N_2$ react with 12 moles of $H_2$ then it gives 8 moles of $NH_3$

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of $N_2$ + Moles of $H_2$

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of $NH_3$

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K$

$V=195.7L$

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V$

where,

w = work done

p = pressure of the gas = 1.0 atm

$\Delta V$ = change in volume = -195.7 L

Now put all the given values in the above formula, we get:

$w=-p\Delta V$

$w=-(1.0atm)\times (-195.7L)$

$w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J$

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, $1.98\times 10^4J$

6 0

3 years ago