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3 years ago

How do the atomic mass and atomic number compare for helium?

Chemistry

2 answers:

ExtremeBDS [4]3 years ago

5 0

The atomic mass is greater!

and that is your answer from me

Sonja [21]3 years ago

4 0

The atomic mass is greater than the atomic number.

You might be interested in

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Nana76 [90]

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

7 0

3 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

5 0

2 years ago

Given 50.0 grams of Strontium, find grams of Phosphorus required for complete reaction

Sveta_85 [38]

Answer:

Mass = 11.78 g of P₄

Explanation:

The balance chemical equation is as follow:

6 Sr + P4 → 2 Sr₃P₂

Step 1: Calculate moles of Sr as;

Moles = Mass / M/Mass

Moles = 50.0 g / 87.62 g/mol

Moles = 0.570 moles

Step 2: Find moles of P₄ as;

According to equation,

6 moles of Sr reacted with = 1 mole of P₄

So,

0.570 moles of Sr will react with = X moles of P₄

Solving for X,

X = 1 mol × 0.570 mol / 6 mol

X = 0.0952 mol of P₄

Step 3: Calculate mass of P₄ as,

Mass = Moles × M.Mass

Mass = 0.0952 mol × 123.89 g/mol

Mass = 11.78 g of P₄

8 0

2 years ago

How many moles of lithium are in 18.2 grams of lithium

Doss [256]

Answer:

I think 2.6 sorry if its wrong

6 0

3 years ago

Read 2 more answers

6.02 kJ>mol. When a small ice cube at -10°C is put into a cup of water at room temperature, which of the following plays a gr

9966 [12]

Answer:

Heat transfer during melting of ice plays greater role in cooling of liquid water.

Explanation:

Temperature of ice = -10 °c

Temperature of water = 0 °c

When ice cube is dipped in to the water.the heat transfer

Q = m c ΔT

⇒ Q = 1 × 2.01 × 10

⇒ Q = 20.1 KJ

Heat transfer during melting of ice $Q_{melt}$ = latent heat of ice

Latent heat of ice = 334 KJ

⇒ $Q_{melt}$ = 334 KJ

Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.

Thus heat transfer during melting of ice plays greater role in cooling of liquid water.

8 0

3 years ago