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oksano4ka [1.4K]
2 years ago
15

What makes the geometry of the tin(4) hydride to have a grater stability?

Chemistry
1 answer:
shtirl [24]2 years ago
4 0
  1. wet seving has ben

Explanation:

1. wet seving has been propsed as amethofolog to study aggregete stability aganist watet erosion

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Will hydrogen ever make a double bond?
Andrei [34K]
No, hydrogen can only hold one bond and that's it.  It only needs to be paired with one bond.  
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3 years ago
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What is the mass number of chlorine?
Anastasy [175]

Answer:

Explanation:

35.453 u

7 0
3 years ago
The number of grams of H2 in 1470 mL of H2 gas. ​
Lorico [155]

Answer:

0.1313 g.

Explanation:

  • It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
  • Suppose that hydrogen behaves ideally and at STP conditions.

<u><em>Using cross multiplication:</em></u>

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

  • Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>

5 0
3 years ago
Which electrophiles can react with an enol?
dlinn [17]

THE MOLECULE HAS A C=C AND AN -OH GROUP, SO IT IS CALLED AN ENE/OL, I.E., AN ENOL. ENOLS CAN BE FORMED ONLY FROM CARBONYL COMPOUNDS WHICH HAVE ALPHA HYDROGENS. THEY CAN BE FORMED BY ACID OR BASE CATALYSIS, AND ONCE FORMED ARE HIGHLY REACTIVE TOWARD ELECTROPHILES, LIKE BROMINE.

8 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
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