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3 years ago

7

What is the mass of two moles of HNO3?

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

126 grams

Explanation:

What is the mass of two moles of HNO3?

H = 1g/mole = 1+

N = 14g/mole = 3-

O = 16g/mole = 2-

HNO3 = 63g/mole

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Which of the following is not a negative consequence of upward urban growth?

Anestetic [448]

Answer:

D. Waste management issues

Explanation:

Answer A: The five major types of pollution - water pollution, air pollution, land pollution, light pollution, noise pollution - have negative effects on the environment, on human health, and our planet as a whole.

Answer B: Urban Heat Island (UHI) is an urban area that is significantly warmer than its surrounding rural areas due to human activities. UHI decreases air quality.

Answer C: The increased use of surrounding land is the main cause of biodiversity loss.

Answer D: Any field of environmental management help people to solve pollution problems. So, D is the correct answer.

....

Hope this answer can help you. Have a nice day!

8 0

2 years ago

17. What is the symbol for the element that forms a.

stepladder [879]

The answer is (C) Al

8 0

3 years ago

A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau

spayn [35]

Answer:

C₅ H₁₂ O

Explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O

6 0

3 years ago

How many grams of solid NaOH are required to prepare a 400ml of a 5N solution? show your work!

Nuetrik [128]

<u>Answer:</u> The mass of solid NaOH required is 80 g

<u>Explanation:</u>

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

$\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}$

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

$\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq$

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

$\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}$

Or,

$\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}$ ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

$5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g$

Hence, the mass of solid NaOH required is 80 g

4 0

3 years ago

What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of

Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 = -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m= molality

molality = $\frac{moles of solute}{mass of solvent in kilograms}$

putting the value in the equation:

molality= $\frac{1.31}{0.53}$

= 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

6 0

3 years ago