Cuales son los símbolos para representar que una reacción se produce en presencia de;

What two things combine with oxygen to form ground-level ozone?

STALIN [3.7K]

Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

-

4 0

2 years ago

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

2 years ago

Which of the following is the method that incorporates designing sustainable living environments with agriculture?

sergey [27]

Answer:

what are the answers?????????

8 0

2 years ago

How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?

lara [203]

<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

Chemistry

Thermodynamics

Specific Heat Formula: q = mcΔT

q is heat (in J)
m is mass (in g)
c is specific heat (in J/g °C)
ΔT is change in temperature (in °C or K)

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

Step 1: Define

Identify variables

[Given] m = 225 g

[Given] c = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] q

Step 2: Solve for q

Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
Multiply: q = (941.4 J/°C)(159.8 °C)
Multiply: q = 150436 J

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0

2 years ago

If you'd like to live in a place that has more moderate summers and winters, it'd be best to pack your suitcases and head to ___

aliya0001 [1]

The answer is Oregon.

6 0

2 years ago

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