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3 years ago

An element A has an atomic number of 11 and another element B has an atomic number 17 (A

Chemistry

1 answer:

Ray Of Light [21]3 years ago

8 0

Answer:

Agroup one.B group seven. lonic/electrovalent bonding.

sodium chloride, 2,8. 2,8,8

Explanation:

the reason why it is an lonicbonding is because it is a force of attraction between opposite charges.it involves the transfer of electrons from a metal to a non-metal.

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A solution of sulfuric acid is titrated with a solution of sodium hydroxide to the phenolphthalein end point. The sample of sulf

qaws [65]

Answer:

a. 2NAOH + H2SO4 —> NA2SO4 + 2H2O

b. The concentration of the acid is 0.0638M

Explanation:Please see attachment for explanation

7 0

3 years ago

How many moles of mgci2 are there in 332 g of the compound

Agata [3.3K]

Find molar mass of MgCl2 so M=<span>95.211 g/mol
N=m/M
N=332/95.2
N=3.487
dont forget to round your sig figs </span>

3 0

3 years ago

Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide accordin

Rudik [331]

Answer:

The limiting reactant is NaOH (option B)

Explanation:

2S(s) + 3O₂(g) + 4NaOH(aq) → 2Na₂SO₄(aq) + 2H₂O(l)

The reaction is ballanced. OK

We need to know how many moles do we have from each compound.

Mass / Molar weight = Mol

Molar weight S = 32 g/m

Molar weight O₂ = 32 g/m

Molar weight NaOH = 40 g/m

Mol S: 2g/ 32g/m = 0.0625 mol

Mol O₂: 3g / 32 g/m = 0.09375 mol

Mol NaOH: 4g/ 40g/m = 0.1 mol

Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.

2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH

0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH

The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.

Let's do the same with O₂

3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH

0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH

5 0

3 years ago

i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t

Ronch [10]

Assuming that the number of mols are constant for both conditions:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Now you plug in the given values. V_1 is the unknown.
$\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}
$
Separate V_1
$V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }$
V= 162.782608696 L

There are 2 sig figs

V= 160 L

3 0

3 years ago

Anyone know this I Need help

aleksandr82 [10.1K]

Answer:

which one?

Explanation:

do you need help on all?

5 0

4 years ago