What are 2 examples of stress the science meaning

8. Ascorbic acid (Vitamin C) cannot be stored in the body, so it must be supplied in the diet. It is a C, H, O compound. If 7.75

spin [16.1K]

Answer:

Molecular formula = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

Explanation:

Step 1: Data given

Mass of vitamin C = 7.75 grams

Mass of CO2 = 11.62 grams

Mass of H2O = 3.17 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 11.62 grams /44.01 g/mol

Moles CO2 = 0.264 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.264 moles CO2 we have 0.264 moles

Step 4: Calculate mass C

Mass C = 0.264 moles * 12.01 g/mol

Mass C = 3.17 grams

Step 5: Calculate moles H2O

Moles H2O = 3.17 grams / 18.02 g/mol

Moles H2O = 0.176 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.176 moles H2O we have 2*0.176 = 0.352 moles

Step 6: Calculate mass H

Mass H = 0.352 moles * 1.01 g/mol

Mass H = 0.356 grams

Step 7: Calculate mass O

Mass O = 7.75 grams - 3.17 - 0.356

Mass O = 4.224 grams

Step 8: Calculate moles O

Mass O = 4.224 grams / 16.0 g/mol

Mass O = 0.264 moles

Step 9: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.264 moles / 0.264 moles = 1

H: 0.352 moles / 0.264 moles = 1.33

O: 0.264 moles / 0.264 moles = 1

This means for each C atom we have 1.33 H atoms and 1 O atom

OR

For every 3 C atoms we have 4 H atoms and 3 O atoms

The empirical formula is C3H4O3

Step 10: Calculate the molecular formula

The molecular mass of the empirical formula is 88.07

Since the molar mass of vitamin C is between 150 and 200 g/mol

We have to multiply the empirical formula by 2

Molecular formula = 2*(C3H4O3) = C6H8O6

The molar mass of vitamin C = 176.14 g/mol

8 0

3 years ago

Calculate the number of grams of carbon dioxide produced from 6 moles of oxygen gas. List only the number of grams of carbon dio

faust18 [17]

Answer:

g CO2(g) = 264.06 g

Explanation:

C(s) + O2(g) → CO2(g)

∴ moles O2(g) = 6 mol

∴ mm CO2 = 44.01 g/mol

⇒ mol CO2(g) = ( 6 mol O2(g) )×( mol CO2(g) / mol O2(g) ) = 6 mol CO2(g)

⇒ mass CO2(g) = ( 6 mol CO2(g) )×( 44.01 g/mol ) = 264.06 g CO2(g)

3 0

4 years ago

CuSO4.5H2O(k) ısı CuSO4(k) + 5 H2O(g) eşitliğine göre bakır sülfatın kristal suyu miktarı ve kaba formülü hesaplanacaktır.

baherus [9]

Answer:

159.609 g/mol

Explanation:

According to the CuSO4.5H2O (k) heat CuSO4 (k) + 5 H2O (g) equation, the crystal water amount of copper sulfate and its rough formula will be calculated.

Weight of copper sulfate containing crystal water = m1 = 249.62… g

Weight of copper sulfate without crystal water weighed = m2 = 159.62 g

Accordingly, calculate the x and y values in the molecular formula of copper sulfate (xCuSO4.yH2O).

3 0

3 years ago

The immediate source of energy that powers a cell’s activities is

Alja [10]

<span>1. The correct option is ADENOSINE TRIPHOSPHATE, ATP. ATP is the basic unit of energy transfer in the living cells. ATP is the principal energy source for metabolic functions, in the cells, ATP are consumed by endothermic metabolic reactions and they are produced by exothermic metabolic reactions.

2. To form ATP, A THIRD PHOSPHATE GROUP HAS TO BE ADDED TO ADP. ADP has two phosphate groups while ATP has three phosphate group. ADP is usulally converted to ATP by the addition of a single phosphate group.

3. ADP and ATP work together and the two can be interconverted. ATP can be hydrolysed to ADP and ADP can be converted to ATP by the addition of a single phosphate group. When energy is needed inside a cell, the ATP will split off one of its phosphate group and become ADP. This split off process produce a high qunatity of energy which is then available for the cell to use.</span>

5 0

3 years ago

Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n

mr_godi [17]

Answer:

V = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

4 0

3 years ago