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3 years ago

The generator is connected to a _, which supplies our _ with electricity.

Chemistry

1 answer:

Kay [80]3 years ago

4 0

Answer:

neutral and current

Explanation:

sorry if im wrong ;(

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The germ theory of disease eventually led to the development of _______.

a_sh-v [17]

The germ theory of disease eventually led to the development of certain types of vaccines. This theory explains that specific type or types of microorganisms, which are bacteria, virus, fungi, or protest species, cause some diseases. These organisms can enter the human body, and can even grow and reproduce, causing infection, inflammation to specific organs in the body, and others. One particular vaccine developed successfully through this understanding is the smallpox vaccine, which is an important vaccine up to this day.

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3 years ago

Read 2 more answers

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

4 0

3 years ago

Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin (c3h8o3) in 135 g of water at 3

pshichka [43]

Data:

Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.

Quesiton: Vapor pressure of water

Solution:

Raoult's Law: The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

Formula: p = Xsolvent * P pure solvent

X solvent = moles solvent / moles of solution

molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol

moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol

molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol

moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol

moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol

Xsolvent = 7.50mol / 7.81mol = 0.960

p = 233.7 torr * 0.960 = 224.4 torr

Answer: 224.4 torr

8 0

4 years ago

For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450

goldfiish [28.3K]

First, assume the order of the given reaction is n, then the rate of reaction i.e. $\frac{dx}{dt}=k\times[A]^{n}$