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3 years ago

K12 3.10 Unit Assessment: Solutions, Part 1 does anyone have the answers for this quiz

Chemistry

2 answers:

Pani-rosa [81]3 years ago

5 0

Eat eat eat eat eat eat eat eat eat

Stells [14]3 years ago

3 0

Answer:

no i was looking for them can you help

Explanation:

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Which is the best way to represent 0.0035 kg by using scientific notation?

Zolol [24]

The answer is 3.5 × 10^-3

7 0

3 years ago

Distinguish between zinc nitrate and lead nitrate

Kobotan [32]

Answer:

Zinc nitrate gives white ppt. which dissolves in excess ammonium hydroxide and produce a colorless solution whereas lead nitrate gives a chalky white ppt. of lead hydroxide which doesnot dissolve.

Explanation:

Hope this helps :)

7 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu

stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = $e^{-2.54738}$

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = $5.61*10^-^5$

6 0

3 years ago

Write out a balanced, molecular equation, total ionic equation, and net ionic equation for each:

dangina [55]

Answer:

1:MgCO3 (s) + 2 NaNO3

2:agcl(s)+kno3(aq)

3:alcl3(aq)+h2(g)

4:NaNO3 + CO2 + H2O

(not sure abt last one)

6 0

3 years ago