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Elden [556K]
3 years ago
8

A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi

s compound?
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal  present in 630 g of the compound is approximately 357 g

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Bradley was working in the lab. He accidentally poured a container of concentrated hydrochloric acid into his container of disti
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Answer is: The solution has now become a good conductor of electricity.

Hydrochloric acid (HCl) dissociate on positive ions or cations of hydogen (H⁺) and negative ions or anions of chlorine (Cl⁻) accordinf to balanced chemical reaction:

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3 years ago
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The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released b
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The complete combustion of propane proceeds through the following reaction:
C_{3} H_{8} + 5O_{2} --> 3CO_{2} + 4H_{2}O

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic. 

The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.

 14.50 kg propane  x <u> 1000 g </u> x <u>  1 mole propane   </u>  x <u>  2220 kJ  </u>   
                                     1 kg              44.1 g                     1 mole

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8 0
3 years ago
Given the balanced chemical equation, SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = −184 kJ Determine the mass (in grams) of H
Ostrovityanka [42]

Answer:

mass HF = 150.05 g

Explanation:

  • SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )

∴ MwHF = 20.0063 g/mol

∴ mol HF = 4 mol

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∴ Q = 345 KJ

mass HF ( mHF ):

⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn

⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ

⇒ mHF = 150.05 g HF

3 0
3 years ago
3.00 L of a gas is collected at 35.0°C and 705.0 mmHg. What is the volume at STP?
makvit [3.9K]

Answer:

2.47L

Explanation:

Using the combined gas law equation as follows:

P1V1/T1= P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 705mmHg

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V1 = 3.00L

V2 = ?

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T2 = 273K (STP)

Using P1V1/T1= P2V2/T2

705 × 3/308 = 760 × V2/273

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Cross multiply

308 × 760V2 = 2115 × 273

234,080V2 = 577,395

V2 = 577,395 ÷ 234,080

V2 = 2.47L

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