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3 years ago

If the atomic number of an element is 6 and its mass number is 15, how many neutrons are in the atom's nucleus?

2 answers:

6 0

Atomic Number is the same as the number of protons in an element.
Mass Number is the number of Protons + Neutrons in an element.

Atomic Number: 6 means 6 Protons
Mass Number: 15 means 15 atoms that are a proton/neutron.
We are given out of the 15 atoms, 6 of them are protons, so the other 9 must be Neutrons.

15 - 6 = 9 so there must be 9 Neutrons.

There are 9 Neutrons in atom's nucleus.

Give Brainiest if you think this is the best answer and explanation. Thanks!

melisa1 [442]3 years ago

5 0

¹⁵₆X
- 6 protons
- 6 electrons
- 15-6 = 9 neutrons

:)

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In a chemical equation, what do the coefficients indicate?

Mekhanik [1.2K]

In a chemical equation coefficients indicate the number of molecules/atoms involved in the reaction.

6 0

3 years ago

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The density of whole milk 1.04 g per mL. What is the volume (in quarts) of 18.5 pounds of whole milk?

nadya68 [22]

The volume : 8,526 quarts

<h3>Further explanation</h3>

Given

The density of whole milk = 1.04 g/ml

mass = 18.5 pounds

Required

The volume

Solution

Conversion of mass

1 pound = 453,592 g

18.5 pounds = 8391,45 g

Density formula:

$\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}$ .

Input the value :

V = m : ρ

V = 8391,45 g : 1.04 g/ml

V = 8068.7 ml

1 ml = 0,00105669 quarts

8068.7 ml =8,526 quarts

3 0

2 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0

3 years ago

formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b

HACTEHA [7]

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

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Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

3 0

3 years ago

Convert 1.248×1010 g to each of the following units.

Leona [35]

Answer:

a) 1.248 x 10⁷ kg

b) 1.248 x 10⁴ Mg

c) 1.248 x 10¹³ mg

d) 1.248 x 10⁴ ton

Explanation:

a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):

1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

8 0

3 years ago