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3 years ago

Which subatomic particle is important in chemical building and helps determine chemical properties

Chemistry

1 answer:

Ann [662]3 years ago

6 0

Explanation:

Since Z is EXPLICITLY determined by the number of protons, massive, positively charged particles, present in the element's nucleus, the proton must be the greatest determinant. Of course, it is the electrons that actually do the chemistry, but Z defines the element.

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if 26.8g of copper (ll) chloride are dissolved in sufficient water to make 4.00 L of solution, what is the molarity of the solut

artcher [175]

Answer: Molarity of the solution is $4.97 \times 10^{-2} M$ and water is the solvent.

Explanation:

Given: Mass of solute = 26.8 g

Volume = 4.00 L

Now, moles of copper (II) chloride (molar mass = 134.45 g/mol) are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{26.8 g}{134.45 g/mol}\\= 0.199 mol$

Molarity is the number of moles of a substance divided by volume of solution in liter.

Therefore, molarity of given solution is calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.199 mol}{4.00 L}\\= 0.04975 M\\= 4.97 \times 10^{-2} M$

Solvent is defined as a component which is present in higher amount in a solution. Generally, a solvent is present in liquid state but it can also be a solid or gas.

In the given solution, copper (II) chloride is dissolved in water so copper (II) chloride is the solute and water is the solvent.

Thus, we can conclude that molarity of the solution is $4.97 \times 10^{-2} M$ and water is the solvent.

4 0

4 years ago

How many atoms are there in 5.10 moles of sulfur?

Alexxx [7]

1 mole ------------- 6.02 x 10²³ atoms
5.10 moles -------- ( atoms sulfur )

atoms sulfur = 5.10 x ( 6.02 x 10²³ ) / 1

atoms sulfur = 3.07 x 10²⁴ / 1

= 3.07 x 10²⁴ atoms of sulfur

hope this helps!

3 0

3 years ago

Read 2 more answers

Given:1 inch = 2.54 cm (exact)1 pound = 454 g1 Liter = 1.0b quarts1 foot = 12 inches1 galion = 4 quarts1 pound = 16 ouncesCalcul

svet-max [94.6K]

We have to convert 0.18 km into feet

First, we need to know the conversions:

1 inch = 2.54 cm (exact)

1 pound = 454 g

1 Liter = 1.0b quarts

1 feet = 12 inches

1 galion = 4 quarts

1 pound = 16 ounces

0.18 km x (1 feet/12 inches) x (1 inch/2.54cm) x (1000 m/1 km) x (100 cm/1 m) = 590 ft

Answer: 590 ft (2 significant figures)

4 0

1 year ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0

3 years ago

Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume

lbvjy [14]

Answer:

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

Molar mass of Glucose(C6H12O6) = 180 g/mol

We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

4 0

4 years ago