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2 years ago

Explain newton's first law of motion

Chemistry

1 answer:

Zina [86]2 years ago

4 0

Answer:

Newton Likes Anime And Hes A Simp he gives all the nee stuff to his crush but she doesn't like him back

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Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati

ollegr [7]

Explanation:

Let us assume that the given data is as follows.

mass of barium acetate = 2.19 g

volume = 150 ml = 0.150 L (as 1 L = 1000 ml)

concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

$Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}$

Hence, moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times Ba(C_{2}H_{3}O_{2})_{2}$ .......... (1)

As, No. of moles = $\frac{mass}{\text{molar mass}}$

Hence, moles of $Ba(C_{2}H_{3}O_{2})_{2}$ will be calculated as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{2.19 g}{255.415 g/mol}$ (molar mass of $Ba(C_{2}H_{3}O_{2})_{2}$ is 255.415 g/mol)

= $8.57 \times 10^{-3}$

Moles of $C_{2}H_{3}O^{-}_{2}$ = $2 \times 8.57 \times 10^{-3}$

= 0.01715 mol

Hence, final molarity will be as follows.

Molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.01715 mol}{0.150 L}$

= 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0

3 years ago

A eudiometer contains a 65.0 ml sample of a gas collected

SCORPION-xisa [38]

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

6 0

3 years ago

ANSWER FAST PLZZZ!!!!!!!!!!!!!

goldfiish [28.3K]

Answer:

1 and 2

Explanation:

1 is negative

2 is negative

3 is positive

5 0

3 years ago

Which of the following is a true statement?

maria [59]

Answer:

the second one seems right :)

Explanation:

3 0

3 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

9 months ago