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balu736 [363]
3 years ago
8

What experiment did J. J. Thomson do?

Chemistry
1 answer:
scoundrel [369]3 years ago
5 0

Answer:

A. He sent a cathode ray between electrically charged metal plates

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A 0.8115 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL
inysia [295]

Answer:

Molarity of NaOH solution is 1.009 M

Explanation:

Molar mass of HCl is 36.46 g/mol

Number moles = (mass)/(molar mass)

So, 0.8115 g of HCl = \frac{0.8115}{36.46}moles HCl = 0.02226 moles HCl

1 mol of NaOH neutralizes 1 mol of HCl.

So, if molarity of NaOH solution is S(M) then moles of NaOH required to reach endpoint is \frac{S\times 22.07}{1000}moles

So, \frac{S\times 22.07}{1000}=0.02226

or, S = 1.009

So, molarity of NaOH solution is 1.009 M

5 0
3 years ago
When we write the formula for an ionic compound, we are merely indicate the "molecules" in the compound with that formula. expla
belka [17]
When writing an ionic compound formula, a "molecular" form is used.  The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻  ⇒ Al₂(SO₄)₃
3 0
3 years ago
How can carbon dating be used to study the history of the earth?
grigory [225]
The amount of carbon in fossils and artifacts decreases when the get older, so you can find out how old an object or fossil is by finding out how much carbon is in it.
7 0
2 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
When combined in the correct ratio hydrogen and oxygen atoms can form water as shown below
Dimas [21]

Answer:molecule of a compound

Explanation:

8 0
2 years ago
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