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3 years ago

The distance between an object and a reference point is the object's __________. displacement neutral point position velocity

Chemistry

1 answer:

sveticcg [70]3 years ago

7 0

Answer:

Displacement

Explanation:

The starting point is different from the end point because the object will have moved a certain distance away. How far the object traveled is the displacement (or change in distance).

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El agua salubre es aquella que tiene más sales disueltas que el agua dulce. En el análisis de una muestra de aguas se encontró q

Slav-nsk [51]

Answer:

0.39 % m/m; 0.42 % m/v; 0.18 % v/v; 4200 ppm; 0.044 mol·L⁻¹; 0.041 mol/kg;

0.089 equiv/L; 0.000 74; 0.999 26

Explanation

Data:

Mass of MgCl₂ = 3.8 g

Volume of solution = 900 mL

Density of solution = 1.09 g/mL

Density of MgCl₂ = 2.32 g/cm³

Calculations

1. Percent m/m

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times 100 \, \%\\\\\text{Total mass} = \text{900 mL} \times \dfrac{\text{1.09 g}}{\text{1 mL}} = \text{981 g}\\\\\text{Mass percent} = \dfrac{\text{3.8 g}}{\text{981 g}} \times 100 \,\% = \textbf{0.39 \% m/m}$

2. Percent m/v

$\text{Mass-by-volume percent} = \dfrac{\text{Mass of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{ Mass-by-volume percent } = \dfrac{\text{3.8 g}}{\text{900 mL}} \times 100 \,\% = \textbf{0.42 \% m/v}$

3. Percent v/v

$\text{Volume percent} = \dfrac{\text{Volume of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{Volume of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mL}}{\text{2.32 g}} = \text{1.64 mL}\\\\\text{ Mass-by-volume percent } = \dfrac{\text{1.64 mL}}{\text{900 mL}} \times 100 \,\% = \textbf{0.18 \% v/v}$

4. Parts per million

$\text{Ppm} = \dfrac{\text{milligrams of solute}}{\text{litres of solution}} = \dfrac{\text{3800 mg}}{\text{0.900 L}} = \textbf{ 4200 ppm}$

5. Molar concentration

$\text{Molar concentration} = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\\text{Moles of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mol}}{\text{95.21 g}} = \text{0.040 mol}\\\\\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.900 L}} = \textbf{0.044 mol/L}$

6. Molal concentration

$\text{Molal concentration} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}$

Mass of water = Mass of solution - mass of solute = 981 g - 3.8 g = 977 g = 0.977 kg

$\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.977 kg}} = \textbf{0.041 mol/kg}$

7. Normality

The normality is the number of equivalents per litre of solution.

The normality of an ion equals the molar concentration times the charge on the ion.

Thus, the normality of MgCl₂ is twice the molar concentration.

Normality = 2 × 0.044 mol·L⁻¹ = 0.089 equiv·L⁻¹

8. Mole fraction of solute

$\chi_{\text{solute}} = \dfrac{n_{\text{solute}}}{n_{\text{total}}}$

Moles of MgCl₂ = 0.040 mol

$\text{Moles of water} = \text{977 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{54.23 mol}$

Total moles = n₂ + n₁ = 0.040 mol + 54.23 mol = 54.27 mol

$\chi_{2} = \dfrac{\text{0.040 mol}}{\text{54.23 mol}} = \mathbf{0.00074}$

9. Mole fraction of solvent

χ₁ = 1 - χ₂ = 1 - 0.000 74 = 0.999 26

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3 years ago

Which graph is an exponential growth model?<br> A<br> B<br> C

gtnhenbr [62]

A the first one represents an exponential growth chart

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3 years ago

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s

Mama L [17]

Use the formula, Q= mcT

Q= heat
m= mass= 1.900Kg= 1.900 x 10^3 grams
c= specific heat= 3.21
T= 4.542 K

Q= (1.900 x10^3g)(3.21)(4.542K)= 14.6 Joules.

7 0

4 years ago

Read 2 more answers

3. What identifies an element?

Anna007 [38]

Answer: Atomic number

Explanation: Atomic numbers never change, they stay the same which is why they are what elements are identified by

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3 years ago

Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g.

AnnZ [28]

Answer:

$m=107.8g$

Explanation:

Hello,

In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h of sleeping:

$E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ$

Then, we compute the energy used that day:

$E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ$

Finally, the mass by considering the consumed fat:

$m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g$

Regards.

4 0

3 years ago