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2 years ago

Please help with this

Chemistry

2 answers:

lord [1]2 years ago

7 0

Choose all options that apply . Which of the following are steps necessary to ensure patient safety ? a) Make sure that the medication is dispensed in the same units in which it was prescribed. b ) Review any calculation questions with the pharmacist . Check to see if there's a better medication for the patient's problem. d) Dispense an extra dose to save the patient from having to return in case of loss or damage to one of the doses. Oe ) Compare the label on the medication with the order from the physician .

dlinn [17]2 years ago

5 0

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Step by Step Explanation?

Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.

The formula is,

P₁ x V₁ = P₂ × V₂

Where,

P₁ is initial pressure = 1 atm

P2 is final pressure = ? (Not Known)

V₁ is initial volume = 10 L

V₂ is final volume = 15 L

Now put the values in the formula,

\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]

Therefore, the answer is 0.67 atm.

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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

Which body changes occur during adolescence?

cupoosta [38]

The answer is C,growth spurts,puberty,& sexual maturity

6 0

3 years ago

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FORENSIC SCIENCE HELP

Annette [7]

That is correct c
Explanation

7 0

3 years ago

What is the electronegativity difference for a bond between potassium and iodine? A. 3.3 B. 4.6 C. 0.4 D. 1.7

user100 [1]

Iodine has an electronegativity of 2.5, and potassium has an electronegativity of 0.8, so the difference is:
2.5 - 0.8 = 1.7

7 0

3 years ago

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What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G

Phantasy [73]

Answer:

the wavelength of radiation emitted is $\mathbf{\lambda= 2169.62 \ nm}$

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

$\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}$

For n = 7:

$\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}$

$\mathtt{E_7 =-0.27755 \ eV}$

For n = 4

$\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}$

$\mathtt{E_4 =- 0.85\ eV}$

The electron goes from the n = 7 to the n = 4, then :

$\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}$

$\mathtt{= 0.57245\ eV}$

Wavelength of the radiation emitted:

$\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}$

where;

hc = 1242 eV.nm

$\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}$

$\mathbf{\lambda= 2169.62 \ nm}$

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3 years ago