All categories

2 years ago

Chemists use letters of the alphabet as ____ for the elements.

Chemistry

1 answer:

d1i1m1o1n [39]2 years ago

8 0

Answer:

The symbol is the right answer.

Explanation:

The “ Symbol” is the correct answer because chemist uses the letters of the alphabet to denote the element. For instance, the element oxygen is denoted by the letter of the alphabet “O”, the hydrogen is denoted by the letter of alphabet “H”, Boron is denoted by the letter of alphabet “B”, etc. Here these are the examples that use one letter but there are other elements that use more than 1 letter as the symbol. For example, the Chlorine is represented by the Cl.

You might be interested in

A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is

vichka [17]

Remember: heat lost = heat gained

When calculating heat loss or gain, remember

mass*(spec heat cap)*(change in T)

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own.

75*x*(96.5-37.1) = 1150*4.184*(37.1-25)

Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

2 years ago

The first eighteen chemical elements on the periodic table are described below in a series of statements. The first 1S letters o

Aleonysh [2.5K]

Answer: The questions looks unclear

Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.

1. D belongs to group 4

E. Belongs to group 7

B belongs to group 1

A belongs to group 8. A noble gas.

R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8

3 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

5 0

2 years ago

Read 2 more answers

The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens

likoan [24]

Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition,
Density = Mass/Volume
= (51.1 g)/(6.63 cm³)
= 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
= 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

4 0

2 years ago