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2 years ago

All isotopes of a given element must have the same

2 answers:

7 0

Isotopes of any given element all contain the same number of protons, so they have the same atomic number (for example, the atomic number of helium is always 2). Isotopes of a given element contain different numbers of neutrons, therefore, different isotopes have different mass numbers.

mafiozo [28]2 years ago

5 0

Answer:

number of protons

Explanation:

Isotopes of any given element all contain the same number of protons, so they have the same atomic number

e.g: the atomic number of helium is always 2.

Isotopes of a given element contain different numbers of neutrons, therefore, different isotopes have different mass numbers.

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3 years ago

For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutraliz

LiRa [457]

Acid-base neutralization reaction is defined as reaction of acid with base to form corresponding salt and water. Strong acid and base completely neutralize each other.

(a) The acid base neutralization reaction is as follows:

$HCl(aq)+NaOH(aq)\rightarrow H_{2}O(l)+NaCl(g)$

From the above balanced chemical reaction, 1 mol of NaOH completely reacts with 1 mol of HCl. The mass of NaOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of NaOH is 39.997 g/mol, thus,

$n=\frac{4.85 g}{39.997 g/mol}=0.121 mol$

Thus, 0.121 mol of NaOH reacts with same amount of HCl and number of moles of HCl will be 0.121 mol.

Since. molar mass of HCl is 36.46 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.121 mol\times 36.46 g/mol=4.421 g$

Therefore, 4.421 g of HCl completely reacts with 4.85 g of NaCl.

(b) The acid base neutralization reaction is as follows:

$2HNO_{3}(aq)+Ca(OH)_{2}(aq)\rightarrow 2H_{2}O(l)+Ca(NO_{3})_{2}(aq)$

From the above balanced chemical reaction, 1 mol of $Ca(OH)_{2}$ completely reacts with 2 mol of $HNO_{3}$ . The mass of $Ca(OH)_{2}$ is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of $Ca(OH)_{2}$ is 74.093 g/mol, thus,

$n=\frac{4.85 g}{74.093 g/mol}=0.06545 mol$

Thus, 0.06545 mol of $Ca(OH)_{2}$ reacts with $2\times 0.06545 mol=0.13091 mol$ of $HNO_{3}$

Since. molar mass of $HNO_{3}$ is 63.01 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.13091 mol\times 63.01 g/mol=8.25 g$

Therefore, 8.25 g of $HNO_{3}$ completely reacts with 4.85 g of $Ca(OH)_{2}$ .

$H_{2}SO_{4}(aq)+2 KOH (aq)\rightarrow 2H_{2}O(l)+K_{2}SO_{4}(aq)$

From the above balanced chemical reaction, 2 mol of KOH completely reacts with 1 mol of $H_{2}SO_{4}$ . The mass of KOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of KOH is 56.1056 g/mol, thus,

$n=\frac{4.85 g}{56.1056 g/mol}=0.0864 mol$

Thus, 0.0864 mol of KOH reacts with $\frac{0.0864 mol}{2}=0.0432 mol$ of $H_{2}SO_{4}$

Since. molar mass of $H_{2}SO_{4}$ is 98.079 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.0432 mol\times 98.079 g/mol=4.24 g$

Therefore, 4.24 g of KOH completely reacts with 4.85 g of $H_{2}SO_{4}$ .

8 0

3 years ago

How many liters of a 3 M NaOH stock solution would you need to make 552 mL of a 105 mM NaOH dilution

Ganezh [65]

Answer:

0.01932 L

Explanation:

First we convert 105 mM to M:

105 mM / 1000 = 0.105 M

Next we convert 552 mL to L:

552 mL / 1000 = 0.552 L

Then we use the following equation:

C₁V₁=C₂V₂

Where:

C₁ = 3 M

V₁ = ?

C₂ = 0.105 M

V₂ = 0.552 L

We input the given data:

3 M * V₁ = 0.105 M * 0.552 L

And solve for V₁:

V₁ = 0.01932 L

7 0

3 years ago