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3 years ago

If an atom has 18 protons, 15 neutrons, and 22 electrons, what is the atom's electrical charge?

Chemistry

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Charge = number of proton - number of electron = 18-22 = -4

SO, OPTION A IS YOUR ANSWER........

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What type of reaction is shown below? Check all that apply. 2Mg+O2—> 2MgO

SSSSS [86.1K]

Answer:

Reaction types are :

Combination(Mg and O combine)
Oxidation (O added to Mg)
Exothermic (light and heat released)

7 0

3 years ago

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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a

nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

= 75.3%

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3 years ago

A intramuscular medication is given at 5.00mg/kg of body weight. What is the dose in grams for a 180-lb patient?

Alex17521 [72]

0.408 gram for 180-lb patient

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3 years ago

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An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0

3 years ago

A closed container holds 2.0 moles of CO2 gas at STP. How many moles of oxygen can be placed in a container of the same size at

PilotLPTM [1.2K]

Answer: 2 moles

Explanation:

STP is Standard Temperature and Pressure. That means the pressure is 1.00 atm and the temperature is 273K. Since the oxygen is placed in the same container, we can use Ideal Gas Law to figure out what container the CO₂ used.

Ideal Gas Law: PV=nRT

P=1.00 atm

n=moles

R=0.08206 Latm/Kmol

T=273K

CO₂

$V=\frac{nRT}{P}$

$V=\frac{(2.0 mol)(0.08206Latm/Kmol)(273K)}{1.00atm}$

$V=44.8L$

Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.

$n=\frac{PV}{RT}$

$n=\frac{(1.00atm)(44.8L)}{(0.08206Latm/Kmol)(273K)}$

$n=1.99=2mol$

There are 2 mol of oxygen.

5 0

3 years ago