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3 years ago

10

How many protons, neutrons and electrons are in a nitrogen atom?

1 answer:

inessss [21]3 years ago

7 0

7 protons , 7 neutrons , 2,5 electrons.

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The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

3 0

2 years ago

What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250

kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

Number of moles = 239 g / 180.2 g/mol

Number of moles = 1.33 mol

Molarity:

Molarity = number of moles / volume in litter

Molarity = 1.33 mol / 0.25 L

Molarity =5.32 M

6 0

3 years ago

1. Given the specific heat of lead is 0.129 J/g.C and that it takes 93.4J of energy to

zheka24 [161]

Answer: 40 grams

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = 93.4J

M = ?

C = 0.129 J/g.C

Φ = 40.4°C - 22.3°C = 18.1°C

Then, Q = MCΦ

Make Mass, M the subject formula

M = Q/CΦ

M = (93.4J) / (0.129 J/g.C x 18.1°C)

M = 93.4J / 2.33J/g

M = 40 g

Thus, the mass of the lead is 40 grams

8 0

3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

2 years ago

There are 300 light particles in an airtight container. Which adjustment would increase the speed of the particles?

tiny-mole [99]

Answer: D

Explanation: Keeping the pressure constant and increasing the temperature.

8 0

2 years ago