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2 years ago

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A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca

rbonate the chemist has added to the flask. Round your answer to 3 significant digits. mmol

1 answer:

svet-max [94.6K]2 years ago

6 0

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}$ .....(1)

Molarity of $BaCl_2$ solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

$1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol$

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

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A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

4 0

2 years ago

You have 125 g of a certain seasoning and are told that it contains 62.0 g of salt. what is the percentage of salt by mass in th

r-ruslan [8.4K]

62.0g/125g= 0.496 x 100 = 49.6%

4 0

2 years ago

Calculate the maximum concentration (in m) of silver ions (ag+) in a solution that contains 0.025 m of co32-. the ksp of ag2co3

Helen [10]

Equilibrium equation is

<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>

<span>From the reaction equation above, the formula for Ksp: </span>

<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>

5 0

2 years ago

Salts and acids are examples of inorganic compounds called _____, which dissociate in water to release ions.

Blizzard [7]

Answer:

Salts and acids are examples of inorganic compounds called <u><em>electrolytes</em></u>.

Explanation:

Electrolytes are the substances which dissociates into ions when dissolved in water and due to this they are able to conduct electric current through them. These compounds in solid form does not conduct electricity due to the absence of free ions.

For example: Sodium chloride , sulfuric acid etc.

$NaCl(aq)\righarrow = Na^+(aq)+Cl^-(aq)$

$H_2SO_4(aq)\rightarrow SO_4^{2-}(aq)+2H^+(aq)$

4 0

2 years ago

Read 2 more answers

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

baherus [9]

Explanation:

Formula for work done is as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now, we will put the given values into the above formula and calculate work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is $3.68 \times 10^{-18} J$ .

7 0

2 years ago