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3 years ago

What’s the answer to the question

Chemistry

2 answers:

san4es73 [151]3 years ago

5 0

Answer: B

Explanation: im pretty sure my teacher mentioned something about N--h bonds are non polar

GalinKa [24]3 years ago

5 0

Answer:

Explanation:

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A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm. (Boyle)

vlabodo [156]

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{4.31 \times 0.755}{1.25} = \frac{3.25405}{1.25} \\ = 2.60324$

We have the final answer as

Hope this helps you

6 0

2 years ago

Select the lewis structure for xeo2f2 which correctly minimizes formal charges.

IrinaVladis [17]

Answer:

explanation and image attached

Explanation:

Our aim is to draw a structure of XeO2F2 whith the least formal charges. We must remember that the compound has 34 valence electrons.

To obtain the least formal charges then Xe must have a total of twelve electrons on its valence shell instead of eight.

The other atoms around the central Xe atom are arranged as shown in the image attached.

Image Credit: UCLA

7 0

3 years ago

When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ______ C to 87 C.

harkovskaia [24]

V1/T1=V2/T2
(15.5)/(360K)=(3.6)/(T2)
T2=83.61290K
T2=-189.3871 degrees Celsius

5 0

3 years ago

Read 2 more answers

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

3 years ago

This is an example of <br> A. Reflection<br> B. Diffraction<br> C. Refraction

lord [1]

Answer:

I believe it is B (if you get it wrong it's my fault)

5 0

3 years ago