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notka56 [123]
2 years ago
14

Pls help will give brainliest

Chemistry
1 answer:
vlada-n [284]2 years ago
4 0

The mass of a present which has a density of 2.15 kg/L and volume of 3650 mL is 7.74 kg.

<h3>How do we calculate mass?</h3>

Mass of any substance will be calculate by using the density and volume by using the below equation as:

Density = Mass / Volume

Given that,

Density of Christmas present = 2.15 kg/L

Volume of present = 3,650 mL = 3.6L

On putting values, we get

Mass = (2.15)(3.6) = 7.74

Hence required mass of Christmas present is 7.74 kg.

To know more about density, visit the below link:
brainly.com/question/10441359

#SPJ1

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An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of the
Katen [24]

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

3 0
3 years ago
____________________ 2. what kind of solution contains more hydrogen ions than hydroxide ions?
JulijaS [17]
The answer is basic solution
HOPE THIS HELPED
8 0
3 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
3 years ago
20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc
bezimeni [28]

Answer:

Concentration of the barium ions  = [Ba^{2+}] = 0.4654 M

Concentration of the chloride ions  = [Cl^{-}]=0.9308 M

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride  = 0.1355 M

n=0.1355 M\times  0.04389 L=0.005947 mol

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

\frac{1}{2}\times 0.05947 mol=0.029735 mol barium hydroxide

Moles of barium hydroxide = 0.029735 mol

Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

=1\times 0.029735 mol= 0.029735 mol

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =[Ba^{2+}]

[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M

Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M

8 0
3 years ago
I’ll give the brainliest to whoever gets it right!!!
kramer

Answer:

itll slowly decay

Explanation:

i hope this helps

8 0
3 years ago
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