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3 years ago

What is the mass number of an atom with 6 protons and 7 neutrons

2 answers:

8 0

Hi, the answer is 13 , because the total of protons and neutrons together form the mass , in other words there are 6 protons and 7 neutrons , 6+7=13.

KiRa [710]3 years ago

6 0

Mass number is 12.01g What you do first to find is find how many protons it such as this atom has 6 so it going to be number 6 on the periodic table then find the mass

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. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.

Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

P1V1 / T1 = P2V2 / T2

-Solve for V2

V2 = P1V1T2 / T1P2

-Substitution

V2 = (6)(1)(252) / (295)(0.45)

- Simplification

V2 = 1512 / 132.75

- Result

V2 = 11.38 L

3 0

3 years ago

The diagram below shows the atoms involved in forming table salt. Which statement best describes what happens next ?

Ghella [55]

An ionic compound is composed of ionic bonds that are formed by transfer of electrons from one atom to the other. The atom that loses electrons acquires a positive charge (cation) while that which gains electrons acquires a negative charge.

In the case of sodium chloride; Sodium Na has 1 electron in its outer orbital while Chlorine Cl has 7 electrons. Thus, Cl requires 1 electron to complete its octet. This electron is donated by Na.

Thus, NaCl is essentially, Na⁺Cl⁻

Ans D) Chlorine becomes an anion by gaining an electron from sodium

5 0

3 years ago

Read 2 more answers

What might be useful for adding solids to a tube

vagabundo [1.1K]

C funnel because the funnel would have a large enough entrance to put a solid through

7 0

4 years ago

a sample of ammonia contains 9g hydrogen and 42g nitrogen. another sample contains 5g hydrogen .calculate the amount of nitrogen

balandron [24]

9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

$9x=42 \cdot 5 \\
9x=210 \\
x \approx 23.33$

The mass of nitrogen in the second sample is 23.33 g.

4 0

4 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago