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3 years ago

Express 6009m in scientific notation

Chemistry

2 answers:

ElenaW [278]3 years ago

8 0

Answer:

The number 6009 in scientific notation is 6.009*10³

Explanation:

Scientific notation is a quick way to represent a number using base ten powers. This notation is used to express very large or very small numbers very easily.

The numbers are written as a product a*10ⁿ where:

a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a number that is not an integer.
n is an integer, which receives the name of exponent or order of magnitude. Represents the number of times the comma moves. It is always an integer, positive if it moves to the left, negative if it moves to the right.

To write the number 6009 in scientific notation, the following steps are performed:

The decimal point moves to the left as many spaces until it reaches the right of the first digit. In this case: 6009=6.009

This number is then written, which will be the coefficient a in the expression of the previous product.
The base 10 is written with the exponent equal to the amount of spaces that the comma moves. This is a positive number because the comma moves to the left. So in this case: the exponent will be 3

Then the number 6009 in scientific notation is 6.009*10³

Andre45 [30]3 years ago

4 0

6.009 x 10^3. You have to put 6009 into a decimal that is less than 10. Count backwards from 9 until you get to a single digit. You would move the decimal back three places making it 10^3

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Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa

Alex777 [14]

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

A sample contains 19.3 g of $N_2$ is reacted with 19.3 g of $F_2$ . Now we need to find the amount of $NF_3$ that can be formed by the complete reactions of each of the reactants.

If all of the $N_2$ was used up in the reaction, how many moles of $NF_3$ would be produced?

Answer: 1.378 moles of $NF_3$ are produced in the reaction.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

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

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, $N_2$ is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of $N_2$ = 19.3 g

Molar mass of $N_2$ = 28.02 g/mol

Putting values in equation 1:

$\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol$

For the given chemical reaction:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

By the stoichiometry of the reaction:

1 mole of $N_2$ produces 2 moles of $NF_3$

So, 0.689 moles of $N_2$ will produce = $\frac{2}{1}\times 0.689=1.378mol$ of $NF_3$

Hence, 1.378 moles of $NF_3$ are produced in the reaction.

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3 years ago

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.

DIA [1.3K]

Answer:

The correct option is: A. 0.168 M

Explanation:

Chemical reaction involved:

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O

Given: For MnO₄⁻ solution-

Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;

For Fe²⁺ solution:

Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M

To find out the concentration of Fe²⁺ solution (M₂), we use the equation:

$\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}$

$\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}$

$0.505 = M_{2}\times 3$

$M_{2} = \frac{0.505}{3} = 0.168M$

Therefore, the concentration or molarity of Fe²⁺ solution: M₂ = 0.168 M

8 0

3 years ago

Calnexin and calreticulin catalyze the removal of the final glucose residue from glycoproteins during the folding process. True

zloy xaker [14]

Answer:

Explanation:

A) False.

Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.

Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.

B) True.

Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.

C) True.

Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.

D) Wrong.

Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.

E) Wrong.

The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.

8 0

3 years ago

Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz

stiv31 [10]

Answer: Freezing point of a solution will be $-1.16^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

$(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}$

$(5.50-T_f)^0C=6.68$

$T_f=-1.16^0C$

Thus the freezing point of a solution will be $-1.16^0C$

3 0

2 years ago

Among the following, which element has the lowest ionization energy? Question 2 options: Cs Na Cl I

marissa [1.9K]

The element with the lowest ionization energy is CESIUM, CS.
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7 0

2 years ago

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