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DiKsa [7]
3 years ago
5

Which six elements are generally considered metalloids

Chemistry
2 answers:
chubhunter [2.5K]3 years ago
4 0

Answer:

boron, silicon, germanium, arsenic, antimony, and tellurium

Eva8 [605]3 years ago
3 0
Boron, silicon, germanium, arsenic, antimony, and tellurium
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What is the mass of CO2 lost at 20 min from the limestone sample?
harkovskaia [24]

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore  

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO​​​ 3

Hence in sample present CaCO​​​ 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO​​​3

8 0
3 years ago
A 25.0g piece of aluminum sits in a room and cools. It loses 4300.0 J of heat. If the initial temperature of aluminum is 125.3°C
Vesna [10]

Answer:

-65.8C

Explanation:

8 0
3 years ago
Problem PageQuestion Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume
IgorC [24]

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.

Be sure your answer has the correct number of significant digits.

Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1 atm  (at STP)

V = Volume of gas = 3.16 L

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K=

n=\frac{PV}{RT}

n=\frac{1atm\times 3.16L}{0.0820 L atm/K mol\times 273K}=0.141moles

CH_4+4Cl_2\rightarrow 4HCl+CCl_4

According to stoichiometry:

4 moles of chlorine produces = 1 mole of carbon tetrachloride

Thus 0.141 moles of methane produces = \frac{1}{4}\times 0.141=0.0352 moles of carbon tetrachloride

volume of carbon tetrachloride =moles\times {\text {Molar volume}}=0.0352mol\times 22.4L/mol=0.788L

Thus volume of carbon tetrachloride that would be produced is 0.788 L

8 0
3 years ago
If 30 ml of water containing 15 mg of salicylate is extracted with 30 ml of chloroform, 2.59 mg of salicylate remains in the wat
In-s [12.5K]

Answer:

Distribution coefficient: 4.79

Explanation:

Distribution coefficient is the ratio between equilibrium concentration of non-aqueous phase and aqueous phase where both solvents are inmiscible. The equation for the problem is:

Distribution coefficient: Concentration in chloroform / Concentration in Water

<em>Concentration in water: 2.59mg / 30mL = 0.08633mg/mL</em>

<em>Concentration in chloroform: (15mg-2.59mg) / 30mL = 0.4137mg/mL</em>

<em />

Distribution coefficient: 0.4137mg/mL / 0.08633mg/mL

<h3>Distribution coefficient: 4.79</h3>
8 0
3 years ago
(a) Calculate the total volume (in liters) of air an adult breathes in a day. (b) In a city with heavy traffic, the air contains
Furkat [3]

Answer:

a) V air/day = 8640 L air  an adult breaths / day

b) 0.0181 L CO intake a person / day

Explanation:

a) one average person has 12 breaths for min:

  • P = 12 breath/min

in each breath it take an average  of 500 mL on air.

  • 1 breath ≅ 500 mL air

⇒ 12 breath / min * 500mL air / breath = 6000 mL air / min

the average air volume per day of a person is:

⇒ Vair/day = 6000 mL air / min * (60 min / h) * ( 24 h / day ) = 8640000 mLair / day * ( L / 1000 mL)

⇒ V air / day = 8640 L / day

b) 2.1 E-6 L CO / L air * 8640 L air / day = 0.0181 L CO / day

4 0
3 years ago
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