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3 years ago

To find the number of neutrons in an atom, subtract its _____.

Chemistry

2 answers:

SpyIntel [72]3 years ago

8 0

Explanation:

It is known that an atom consists of protons, neutrons and electrons. Both protons and neutrons are located inside the nucleus of an atom whereas electrons revolve around the nucleus.

Atomic number is the total number of protons present in an atom. On the other hand, atomic mass is the sum of total number of protons and neutrons present in an atom.

That is, Atomic mass = no. of protons + no. of neutrons

no. of neutrons = atomic mass - no. of protons

Therefore, we can conclude that to find the number of neutrons in an atom, subtract its number of protons from atomic mass.

Komok [63]3 years ago

6 0

Subtract it's atomic number from it's mass number

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Name 2 gram negative bacteria often found in contaminated food

Alex17521 [72]

E. coli and Salmonella spp.

6 0

2 years ago

1. Using the balanced equation, answer the following questions:

nalin [4]

Answer:

a) 2.53 × 10²³ molecules of O₂

b) 31.90 g of KCl

Explanation:

The balance chemical equation for given decomposition reaction is as follow;

2 KClO₃ → 2 KCl + 3 O₂

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

Moles = Mass / M.Mass

Moles = 34.35 g / 122.55 g/mol

Moles = 0.280 moles of KClO₃

Step 2: <u>Find out moles of O₂ produced;</u>

According to eq,

2 moles of KClO₃ produces = 3 moles of O₂

So,

0.280 moles of KClO₃ will produce = X moles of O₂

Solving for X,

X = 0.280 mol × 3 mol / 2 mol

X = 0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules = Moles × 6.022 × 10²³

No. of Molecules = 0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules = 2.53 × 10²³ molecules of O₂

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

Moles = Mass / M.Mass

Moles = 52.53 g / 122.55 g/mol

Moles = 0.428 moles of KClO₃

Step 2: <u>Find out moles of KCl produced;</u>

According to eq,

2 moles of KClO₃ produces = 2 moles of KCl

So,

0.428 moles of KClO₃ will produce = X moles of KCl

Solving for X,

X = 0.428 mol × 2 mol / 2 mol

X = 0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

Mass = Moles × M.Mass

Mass = 0.428 mol × 74.55 g/mol

Mass = 31.90 g of KCl

6 0

2 years ago

(5.0 m3) ( 7.5 mmHg) = (P)(4.0m3)

anzhelika [568]

It looks like we are solving for a pressure. All that is required is some algebraic manipulation to find our pressure in mmHg.

Given:

(5.0 m³)(7.5 mmHg) = (P)(4.0m³)

Multiply:

37.5 = 4.0P

Divide:

9.375 = P

P = 9.4 mmHg (remember sig figs)

<h3>Answer:</h3>

9.4 mmHg

7 0

2 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0

3 years ago

g How many turns through the citric acid cycle are required to fully oxidize all of the acetly-coA that result from 1 molecule o

faust18 [17]

Answer:

The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.

Explanation:

Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:

Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺

Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.

2 Pyruvate ----> 2 AcetylCoA + 2CO₂

Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.

7 0

3 years ago