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3 years ago

5

How do you express a number in scientific notation?

1 answer:

kykrilka [37]3 years ago

5 0

D. scientific notation is (insert decimal number) x10^(however many 0's there are)

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Would you rather be hot or cold

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cold

Explanation:

because being cold helps to refresh my body personally

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2 years ago

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A _________________ is the path of energy transfer from producer to consumers.

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2 years ago

Define mechanical energy. Using the example of the pendulum, explain how kinetic and potential energy relate to the mechanical e

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2 years ago

How manymoles of each ion are present in 175 mL of 0.147 M Fe2(SO4)3?

lilavasa [31]

Answer: 0.0257 moles of $Fe^{3+}$ and 0.0257 moles of $SO_4^{2-}$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{moles}{\text {Volume in L}}$

moles of $Fe_2(SO_4)_3=Molarity\times {\text {Volume in L}}=0.147\times 0.175L=0.0257moles$

The balanced reaction for dissociation will be:

$Fe_2(SO_4)_3\rightarrow Fe^{3+}+SO_4^{2-}$

According to stoichiometry:

1 mole of $Fe_2(SO_4)_3$ gives 1 mole of $Fe^{3+}$ and 1 mole of $SO_4^{2-}$

Thus there will be 0.0257 moles of $Fe^{3+}$ and 0.0257 moles of $SO_4^{2-}$

8 0

3 years ago

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

2 years ago