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2 years ago

What are the Valence Electrons of

Chemistry

1 answer:

ollegr [7]2 years ago

5 0

Answer:

Hidrogen - 3.04

Lithium - .98

Magnesium - 1.31

Beryllium - 1.57

Sodium - .93

Helium - none

Argon - none

Neon - none

Boron - 2.04

Silicon - 1.9

Potassium - 1

Oxygen - 3.44

Chlorine - 3.16

Fluorine - 3.98

Nitrogen - 3.04

Phophorus - 2.19

Aluminum - 1.61

Carbon - 2.55

Calcium - 1.36

Explanation: hope it helps

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If the total mass of the products of a reaction is 90 gram, what was the total mass of the reactants

Marizza181 [45]

90 grams due to laws of conservation of mass. Output mass = input mass. Mass can never be created or destroyed.

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3 years ago

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3 years ago

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The amount ofcalcium present in milk can be determined by adding oxalate to asample and measuring the massof calcium oxalate pre

Sauron [17]

<u>Answer:</u> The mass percent of calcium in milk is 0.107 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of calcium oxalate = 0.429 g

Molar mass of calcium oxalate = 128.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium oxalate}=\frac{0.429g}{128.1g/mol}=0.0033mol$

The given chemical equation follows:

$Na_2C_2O_4(aq.)+Ca^{2+}(aq.)\rightarrow CaC_2O_4(s)+2Na^+(aq.)$

Sodium oxalate is present in excess. So, it is considered as an excess reagent. And, calcium ion is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium oxalate is produced from 1 mole of calcium ion

So, 0.0033 moles of calcium oxalate is produced from = $\frac{1}{1}\times 0.0033=0.0033mol$ of calcium ions

Now, calculating the mass of calcium ions by using equation 1, we get:

Moles of calcium ions = 0.0033 moles

Molar mass of calcium ions = 40 g/mol

Putting values in equation 1, we get:

$0.0033mol=\frac{\text{Mass of calcium ions}}{40g/mol}\\\\\text{Mass of calcium ions}=(0.0033mol\times 40g/mol)=0.132g$

To calculate the mass percentage of calcium ions in milk, we use the equation:

$\text{Mass percent of calcium ions}=\frac{\text{Mass of calcium ions}}{\text{Mass of milk}}\times 100$

Mass of milk = 125 g

Mass of calcium ions = 0.132 g

Putting values in above equation, we get:

$\text{Mass percent of calcium ions}=\frac{0.132g}{125g}\times 100=0.107\%$

Hence, the mass percent of calcium in milk is 0.107 %

7 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

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3 years ago

A drop of water (H20) has a mass of 2.34 x 10e-2 grams. How many

creativ13 [48]

Answer:

2333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

Explanation:

2x3=1000

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3 years ago