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4 years ago

What is the reading on your balance and that of your partner?what do these values represent?

1 answer:

3 0

These values may represent the amount of pulling force that we exert on each other or in other words exert forces on both parties.

The answer may differ as well but keep in mind the values should be equal.

Note:

The answer is not about helping each other or the "spring balance". We are not talking about the literal moral values instead we are talking about Chemistry and Science.

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Please answer this Q:

Anarel [89]

D. container four because of the water

3 0

3 years ago

How many milliliters of water are needed to prepare a 3.5M solution of NaOH if you have .5mol of the solute

Mamont248 [21]

Answer:

1335.12 mL of H2O

Explanation:

To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).

From the molarity formula we can first calculate the volume of the solution:

$M=\frac{solute moles}{solution volume}$

$Solutionvolume=\frac{solute moles}{M} =\frac{5mol}{3.5\frac{mol}{L} } =1.429L$

The volume of the solution as we said previously is:

Solution volume = solute volume + solvent volume

To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:

$MW=\frac{mass}{mol}$

$Mass=MW*mol=39.997\frac{g}{mol} *5mol=199.985g$

Now with the density of NaOH the milliliters of solute can be determined:

$d=\frac{mass}{volume}$

$Volume=\frac{mass}{d} =\frac{199.985g}{2.13\frac{g}{mL} } =93.88mL of NaOH$

Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:

Solvent volume = solution volume - solute volume

Solvent volume = 1429 mL - 93.88 mL = 1335.12 mL of H2O

7 0

4 years ago

A 2.50 L balloon is filled with water at 2.27 atm. If the balloon is squeezed into a 0.80 L beaker and does NOT burst, what is t

Ludmilka [50]

Answer:

$P_2=7.09atm$

Explanation:

Hello!

In this case, given the change in volume and pressure of the gas, it is possible for us to recall the Boyle's law as way to understand the inversely proportional relationship between pressure and volume:

$P_1V_1=P_2V_2$

Thus, when solving for the final pressure, P2, given the initial pressure and volume and the final volume, we obtain:

$P_2=\frac{P_1V_1}{V_2}\\\\P_2=\frac{2.27atm*2.50L}{0.80L}\\\\P_2=7.09atm$

Best regards!

3 0

3 years ago

Read 2 more answers

What is the missing number in the nuclear reaction shown below 175 78 pt?

Dmitry [639]

Above question is incomplete. Complete question is
<span>What is the missing number in the nuclear reaction 175 78 pt -- 4 2 He plus 76 Os?
....................................................................................................................
Solution:

Reaction involved in present nuclear event is
</span>175 78 Pt → 4 2 He + 76 Os

In above nuclear reaction, atomic mass number of Os is missing. The atomic mass number of Os = 175 - 4 = 171

Thus, complete nuclear can be written as
175 78 Pt → 4 2 He + 171 76 Os

3 0

3 years ago

Read 2 more answers

One of relatively few reactions that takes place directly between two solids at room temperature is Ba(OH)2 · 8H2O + ammonium th

Hatshy [7]

Answer:

$\boxed{\text{3.1 g}}$

Explanation:

We will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 315.46 76.12

Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O

m/g: 6.5

1. Moles of Ba(OH)₂·8H₂O

$\text{Moles of Ba(OH)$_{2}\cdot $8H$_{2}$O}\\= \text{ 6.5 g Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}{\text{ 315.46 g Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}$

2. Moles of NH₄SCN

The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

$\text{Moles of NH$_{4}$SCN} =\text{0.0206 mol Ba(OH)$_{2}\cdot $8H$_{2}$O} \times \dfrac{\text{2 mol NH$_{4}$SCN}}{\text{1 mol Ba(OH)$_{2}\cdot $8H$_{2}$O}}\\= \text{0.0412 mol NH$_{4}$SCN}$

3. Mass of NH₄SCN

$\text{Mass of NH$_{4}$SCN} = \text{0.0412 mol NH$_{4}$SCN} \times \dfrac{\text{76.12 g NH$_{4}$SCN}}{\text{1 mol NH$_{4}$SCN}} =\\\textbf{3.1 g NH$_{4}$SCN}\\\\\text{You must use }\boxed{\textbf{3.1 g}}\text{ NH$_{4}$SCN}$

5 0

4 years ago