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fgiga [73]
3 years ago
6

Plz help me i swear will mark you brainiest :(

Chemistry
1 answer:
vladimir1956 [14]3 years ago
7 0

Answer:

C. the relative molecular mass of the compound

Explanation:

Like molecular formulas, empirical formulas are not unique and can describe a number of different chemical structures or isomers. <u>To determine an empirical formula, the relative molecular mass of the composition of its elements</u> can be used to mathematically determine their ratio.

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What is the compound SCI2
lys-0071 [83]

Answer:

It is called Sulfur dichloride

3 0
3 years ago
2. You have 200g or a solution that contains 30g of hydrochloric acid (HCI),
alex41 [277]

Answer:

the percentage of your solution that made up of HCI acid is 15%

Explanation:

The computation of the percentage of your solution that made up of HCI acid is given below:

Given that

There is 200g or a solution that have 0 g of hydrochloric acid (HCI)

Based on the above information

The percentage is

= 30g ÷ 200g

= 15%

Hence, the percentage of your solution that made up of HCI acid is 15%

7 0
3 years ago
I have 2 questions....but it's multiple choice for 1 of them.
Taya2010 [7]

1. Blood is an organic compound

2. B)

4 0
3 years ago
Read 2 more answers
A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant
KiRa [710]

Answer:

V_2=2.995L\\\\W=248.5J

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

P_1V_1=P_2V_2

So we solve for V_2 by firstly computing the initial pressure:

P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L}  =11.98atm

V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L

Finally, we can compute the work by using the following formula:

W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J

Best regards.

4 0
3 years ago
Someone explain it plz​
timofeeve [1]

Answer:

1). 1 mole of Carbon burnt in air

C + O2 →CO2

1 mole of carbon produces 1 mole of CO2 which is 44g of CO2

2). 1 mole of carbon is burnt in 16g of dioxygen

32g of O2 = 44g of CO2

1g of O2 = 44/32

CO2 (Dioxygen is limiting reagent)

16g of O2 = 4/32 × 16 = 22g of CO2 in one mole

3) 2 moles of Carbon burnt in 16g of dioxygen

16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2

4 0
2 years ago
Read 2 more answers
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