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3 years ago

Describe several products of photosynthesis important to humans

1 answer:

6 0

The products of photosynthesis which are important to humans are OXYGEN AND GLUCOSE. The main product of photosynthesis is GLUCOSE ,which is the molecule that produces energy to run the process of the cell.six molecules of carbondioxide and six molecules of water are needed to produce one molecule of glucose. During the process of photosynthesis,carbondioxide and water combines in the presence of sunlight and chlorophyll to produce carbohydrates and oxygen. Oxygen is essential for breathing in humans.

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What two factors most affect the pitch of the sounds made by a person's vocal cords? A. the volume and health of the person's lu

dusya [7]

Answer:

The pitch of the sound a person makes is determined by several factors, including the size and tension of his or her vocal cords.

Explanation:

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3 years ago

Which of the following would be a result of increased solar activity?

scoundrel [369]

The question is asking us "Which of the following would be a result of increased solar activity?" The Sun is extremely hot, and when it is warm on Earth, the heat has as it source in the Sun. Therefore, an increased solar activity would mean increased temperature on Earth - longer summers and shorter winters, and warmer sea temperatures. The best answer is : B. Warmer-than-normal sea surface temperatures

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3 years ago

Read 2 more answers

A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.

makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> smallest whole number ratio of elements in a compound

Molecular Formula => actual whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

% => grams => moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H => 7.14 : 14.29

To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0

3 years ago

Please please help:(

borishaifa [10]

Magnesium and silver

4 0

3 years ago

The amount ofcalcium present in milk can be determined by adding oxalate to asample and measuring the massof calcium oxalate pre

Sauron [17]

Answer: The mass percent of calcium in milk is 0.107 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of calcium oxalate = 0.429 g

Molar mass of calcium oxalate = 128.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium oxalate}=\frac{0.429g}{128.1g/mol}=0.0033mol$

The given chemical equation follows:

$Na_2C_2O_4(aq.)+Ca^{2+}(aq.)\rightarrow CaC_2O_4(s)+2Na^+(aq.)$

Sodium oxalate is present in excess. So, it is considered as an excess reagent. And, calcium ion is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium oxalate is produced from 1 mole of calcium ion

So, 0.0033 moles of calcium oxalate is produced from = $\frac{1}{1}\times 0.0033=0.0033mol$ of calcium ions

Now, calculating the mass of calcium ions by using equation 1, we get:

Moles of calcium ions = 0.0033 moles

Molar mass of calcium ions = 40 g/mol

Putting values in equation 1, we get:

$0.0033mol=\frac{\text{Mass of calcium ions}}{40g/mol}\\\\\text{Mass of calcium ions}=(0.0033mol\times 40g/mol)=0.132g$

To calculate the mass percentage of calcium ions in milk, we use the equation:

$\text{Mass percent of calcium ions}=\frac{\text{Mass of calcium ions}}{\text{Mass of milk}}\times 100$

Mass of milk = 125 g

Mass of calcium ions = 0.132 g

Putting values in above equation, we get:

$\text{Mass percent of calcium ions}=\frac{0.132g}{125g}\times 100=0.107\%$

Hence, the mass percent of calcium in milk is 0.107 %

7 0

3 years ago