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2 years ago

I NEED HELP ASAP !!!!!!!!!!!!!!!!!!!!!1

Chemistry

1 answer:

Karolina [17]2 years ago

6 0

Answer:

true im pretttyyy sure

Explanation:

bcuz the stronger the intermolecular forces the higher the boiling point :3

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Answer this please thank you

jekas [21]

Answer:

Heterozygous

Explanation:

4 0

3 years ago

Hi, Can someone help me with this question, please! It's due tonight!

-BARSIC- [3]

Answer:

24.0 g C₃H₈

Explanation:

To find the mass of C₃H₈, you need to (1) convert grams CO/H₂ to moles CO/H₂ (via molar mass), then (2) convert moles CO/H₂ to moles C₃H₈ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles C₃H₈ to grams C₃H₈ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs in the given values.

Molar Mass (CO): 12.011 g/mol + 15.998 g/mol

Molar Mass (CO): 28.009 g/mol

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (C₃H₈): 3(12.011 g/mol) + 8(1.008 g/mol)

Molar Mass (C₃H₈): 44.097 g/mol

3 CO + 7 H₂ ----> 1 C₃H₈ + 3 H₂O
^ ^ ^

45.8 g CO 1 mole 1 mole C₃H₈ 44.097 g
----------------- x ------------------ x -------------------- x ------------------ =
28.009 g 3 moles CO 1 mole

= 24.0 g C₃H₈

87.3 g H₂ 1 mole 1 mole C₃H₈ 44.097 g
---------------- x --------------- x --------------------- x ----------------- =
2.016 g 7 moles H₂ 1 mole

= 273 g C₃H₈

It was necessary to find the mass of the products from both of the reactants because you did not know which one was the limiting reagent. The limiting reagent is the reactant which is completely used up first. Because CO produced the smaller amount of product, it must be the limiting reagent. Therefore, the actual amount of C₃H₈ produced is 24.0 grams.

4 0

2 years ago

How did Thomson measure the charge to mass ratio of the electron? a.He shot helium nuclei into gold foil to measure the nuclei s

Ket [755]

Answer:

The correct answer is option b. "He passed cathode rays through a magnetic field and measured the deflection".

Explanation:

J. J. Thomson was a notorious scientist that did not only made the discovery of the electron, but he measure its charge to mass ratio. In order to determine this property, Thomson passed cathode rays through a magnetic field and measured the deflection. Cathode rays are actually composed of particles, which are now known as electrons, and its charge to mass ratio is about 10^8 coulomb per gram.

3 0

3 years ago

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Nana76 [90]

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

7 0

3 years ago

A+gas+at+250k+and+15atm+has+a+molar+volume+12%+less+than+the+perfect+gas+law+character. +calculate+the+compressibility+factor

shutvik [7]

The compression factor based on the information given is 0.083.

<h3>How to calculate the compression factor?</h3>

The way to calculate the compression factor goes this:

V(ideal) = RT/P

= (0.0821 × 250)/15

= 1.368

V(real) will be:

= V(ideal)/12

= 1.368/12 = 0.114

The compression factor will be:

= 1.368/(12 × 1.368)

= 0.083

Therefore, the compression factor is 0.083.

Learn more about compression factor on:

brainly.com/question/24261456

#SPJ12

6 0

2 years ago