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asambeis [7]
2 years ago
5

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

Chemistry
2 answers:
ira [324]2 years ago
6 0
No se perdón que no pueda ayudar:(
irina [24]2 years ago
4 0

Answer:

gfchjklnmbvh

Explanation:

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7. Calculate the net charge of the following atoms.
loris [4]

Answer:

any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have.

Explanation:

6 0
2 years ago
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What is 25 m rounded to one significant figure
den301095 [7]

Rounded to 1 significant figure, 25 m would go to 30. This is because 0 isn't significant, so the 3 is the only significant figure.

8 0
3 years ago
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Explain how the presence of several oxygen atoms ina compound containing an-OH group can make the compound acidic.
Lina20 [59]
<span>In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X"
</span>

The factors to consider

Two of the factors which influence the ionisation of an acid are:

<span>the strength of the bond being broken,the stability of the ions being formed.</span>

In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar.

7 0
3 years ago
Which adaptation is likely to increase the chances of survival of an animal in a rainforest?
LiRa [457]
I know I think it’s a A

6 0
3 years ago
Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting t
Masja [62]

Answer:

Molecular formula of hydrocarbon is: C₂H₆

Explanation:

The combusting of the hydrocarbon:

CxHy + O₂ → CO₂ + H₂O

Using gas law to obtain molar mass of the gas mixture (CO₂ + H₂O):

P/RT = n/V

Where:

P is pressure (1,208 atm)

R is gas constant (0,082 atmL/molK)

T is temperature (375 K)

n/V = 0,0393 mol/L

1,1128 g/L ÷ 0,0393 mol/L = 28,32 g/mol

Thus, average molecular weight is:

28,32 g/mol = 44,01 g/mol X + 18,02 g/mol Y

1 = X + Y

Where X is CO₂ molar percentage and Y is H₂O molar percentage.

Solving:

X = 0,397

Y = 0,603

39,7% H₂O

60,3% CO₂

With this proportion you can obtain ratio CO₂:H₂O thus:

60,3/39,7 = 1,52 So, 2 CO₂: 3H₂O

The moles of the hydrocarbon are:

PV/RT = n

P (0,959 atm)

V ( 1/5 of final volume)

T (298K)

n = 7,85x10⁻³ mol

The moles of CO₂ are:

0,0393 mol × 2 mol CO₂/ 5 mol total =0,01572.

Ratio of CO₂:CxHY =

0,01572 : 7,85x10⁻³ 2 CO₂: 1 CxHY

Doing:

1 CxHy + O₂ → 2 CO₂ + 3 H₂O

By mass balance:

1 C₂H₆ + 7/2 O₂  → 2 CO₂ + 3 H₂O

Thus, molecular formula of hydrocarbon is: <em>C₂H₆ </em>

I hope it helps!

7 0
3 years ago
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