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Naya [18.7K]
2 years ago
15

Nah I'm gonna fail this quiz help please ​

Chemistry
1 answer:
Eva8 [605]2 years ago
4 0

Answer:

It has 2 eyes

Explanation:

Quantitive observation means that it is facts based of numbers, therefore you can count how many eyes it has, but not that it has red eyes

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A gas occupies the volume of 215ml at 15C and 86.4kPa?
kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L

Hope this helps!

3 0
4 years ago
A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
sertanlavr [38]

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

4 0
3 years ago
Write the word equation for the chemical reaction: Zn(s)+2HCl2(aq)--->ZnCl2(aq)+H2(g)
Vikki [24]
Zn(s) + 2HCl(aq) = ZnCl₂(aq) + H₂(g)

zinc + hydrochloric acid = zinc chloride + hydrogen
4 0
3 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
In humans, the allele for curly hair and the allele for straight hair show incomplete dominance the offspring of a curly haired
Vika [28.1K]
A
eplanation
i meannn they both had curly hair
3 0
3 years ago
Read 2 more answers
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