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Levart [38]
3 years ago
5

If an element gives away an electron, will it form a positive ion or a negative ion?

Chemistry
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

posotive

Explanation:

electrons give off negative pulse so take one away

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
Is o-toluic acid soluble in ether, NaOH?, Is Napthalene soluble in NaOH?
natima [27]
Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.

Naphthalene is insoluble in NaOH.
8 0
3 years ago
A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p
timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0
3 years ago
I need help with this ASAP
NNADVOKAT [17]

Answer: I dunno that but u should give me brainliest

Explanation:

5 0
2 years ago
How many atoms of hydrogen in the product of the equation below:
Luden [163]

Answer: 12 atoms

Explanation: the equation has 6H2 i.e

6*2 = 12

6 0
3 years ago
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