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ivanzaharov [21]
3 years ago
6

How many atoms are in a casein molecule?

Chemistry
1 answer:
konstantin123 [22]3 years ago
4 0
The heavy atom count of a casein molecule is 143 I believe
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A salt contains only magnesium and one of the halide ions. A 0.0776-g sample of the salt was dissolved in water, and an excess o
sdas [7]

Answer:

The formula of the magnesium halide is MgF₂

Explanation:

All halides, X, produce a salt with Mg with the formula:

MgX₂

<em>-There are 2 moles of the halide ion per mole of Mg-</em>

<em />

With the mass of the MgSO₄ we can find moles of magnesium sulfate  = Moles Mg.

With moles of Mg we can know the moles of the halide -1 mole Mg = 2 moles of Halide-

And we can find the mass of Mg in the 0.0776g sample. Subtracting we can find the mass of the halide and, with the mass and moles of the halide we can find its molecular weight and its identity:

<em>Moles MgSO₄ -Molar mass: 120.366g/mol- = Moles Mg:</em>

0.150g * (1mol / 120.366g) = 1.2462x10⁻³ moles Mg

<em>Moles halide:</em>

1.2462x10⁻³ moles Mg * 2 = <em>2.4924x10⁻³ moles Halide</em>

<em>Mass Mg -Molar mass: 24.305g/mol:</em>

1.2462x10⁻³ moles Mg * (24.305g / mol) = 0.0303g Mg

<em>Mass halide:</em>

0.0776g - 0.0303g Mg = <em>0.0473g</em>

<em>Molecular weight of the halide:</em>

0.0473g / 2.4924x10⁻³ moles =

18.98g/mol

This molecular weight is the molecular weight of Fluoride ion, F⁻,

<h3>The formula of the magnesium halide is MgF₂</h3>
6 0
3 years ago
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
What is (part of life)and(art of life)?someone will answer me​
irinina [24]

Explanation:

Part of life is the role we certainly play during our role in life.Like a quote of shakespeare,we have our roles in earth to perform or we are like a chess whom we don't want to waste a move.

Art of life signifies that our life is wonderful with mysteries and hypotenic beauty and wonders and with different moments in a journey.

<em>Keep</em><em> </em><em>smiling </em><em>and</em><em> </em><em>hope</em><em> </em><em>u</em><em> </em><em>r</em><em> </em><em>satisfied </em><em>with</em><em> </em><em>my</em><em> </em><em>answer</em><em>.</em><em>Have</em><em> </em><em>a</em><em> </em><em>great</em><em> </em><em>day</em><em> </em><em>:</em><em>)</em>

7 0
2 years ago
4. What electrons are involved in chemical bonding?
Yanka [14]

Answer:

The electrons that participate in chemical bonds are the valence electrons, which are the electrons found in an atom 's outermost shell.

Explanation:

4 0
3 years ago
A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the
Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0
1 year ago
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