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3 years ago

Which method of galaxy formation is expected to dominate far in the future? (1 point

Chemistry

1 answer:

nevsk [136]3 years ago

6 0

Answer:

Dark Matter

Explanation:

90 percent of the mass in the universe is dark matter

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Find the relative molecular mass of hydrated iron (II) sulphate FeSO4.7H2O

Pepsi [2]

Answer:

the relative molecular mass of hydrated iron (II) sulfate FeSO4.7H2O is 278.02

Explanation:

5 0

3 years ago

You set up an experiment to determine the conditions needed to form the gemstones with the most clarity. You gather data from 5

Andreyy89

I’m in chemistry too. Giving this my best guess I’d say
Independent: Gem Mines
Dependent: Gem clarity
Controlled: Temperature

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

3 years ago

A solution is prepared by dissolving 51 g of salt in 340 g of

garri49 [273]

Answer:

okķkkkkkkkkkkkkkkkkkkkkk.I don't know the answer. GIVE THANKS

5 0

3 years ago

If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoret

Mariana [72]

The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.

0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield

3 0

3 years ago