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antoniya [11.8K]
3 years ago
14

What is the composition of the supercluster

Chemistry
1 answer:
sergey [27]3 years ago
7 0
Trick question due to the fact that the component clusters are generally shifting away from each other due to the Hubble flow
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A sample of strontium bicarbonate weighs 5.0020 mg. How many oxygen atoms are in the sample? I have the answer, but don't unders
hodyreva [135]

There are 14.4 * 10^18 oxygen atoms in 5.0020 mg of  strontium bicarbonate.

Mass of strontium bicarbonate Sr(HCO3)2 = 5.0020 mg

Molar mass of strontium bicarbonate Sr(HCO3)2 = 209.6537 g/mol

Number of moles of strontium bicarbonate Sr(HCO3)2 = 5.0020 * 10^-3g/209.6537 g/mol

= 2.3858 * 10^-5 moles

Given that we have 6 oxygen atoms per molecule of Sr(HCO3)2, the total number of oxygen atoms in Sr(HCO3)2 becomes;

2.3858 * 10^-5 moles * 6.02 * 10^23 = 14.4 * 10^18 oxygen atoms

Learn more: brainly.com/question/9743981

5 0
3 years ago
Can someone please help? This is really hard!!
Arte-miy333 [17]

Answer:

Less

Explanation:

The hydronium from the HCl is used to neutralize the bicarbonate in the baking soda. The hydronium is the acid and the bicarbonate ion is the base while the sodium and chloride ions are pH-neutral. Since theres a 1:1 mole ratio of hydronium to HCl and bicarbonate to sodium bicarbonate, it would require less HCl to neutralize a less concentration baking soda solution.

6 0
2 years ago
URGENT!!!!!
Nikitich [7]

Answer:

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6 0
3 years ago
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
How many grams are in 1 mole of Pb3(PO4)4?
kvasek [131]

Answer:

In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams

Explanation:

This compound is the lead (IV) phosphate.

Grams that occupy 1 mole, means the molar mass of the compound

Pb = 207.2  .3 = 621.6 g/m

P = 30.97 .4 = 123.88 g/m

O = (16 .  4) . 4 = 256 g/m

621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m

6 0
3 years ago
Read 2 more answers
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