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2 years ago

Calculate the mass in grams of 1.32x10^20 uranium atoms

1 answer:

6 0

Uranium has an atomic weight of 238.03 g/mol. We have 1.32x10^20 atoms of uranium. We must convert this to moles by dividing Avogadro's number 6.022x10^23 atoms/mol. (1.32x10^20atoms)/(6.022x10^23atoms/mol) 2.19x10^-4moles of Uranium Now multiply this by the atomic weight of Uranium. 2.19x10^-4mol*238.03g/mol Grams of Uranium = .0522g

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In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all

Brilliant_brown [7]

Full Question:

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air

boiling point of –1.1°C

odorless

does not react with water

burns readily in air

does not react with water

<h3>Explanation:</h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

6 0

2 years ago

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

Dennis_Churaev [7]

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

Volume of the calcium hydroxide solution used is 0.0235 mL.

4 0

2 years ago

A 1.0 x 102- gram sample is found to be pure alanine, an amino acid found in proteins. How many moles of alanine are in the samp

Romashka-Z-Leto [24]

Answer:

1.123x10⁻⁴ moles of alanine

Explanation:

In order to convert grams of alanine into moles, we need to know its molecular weight:

The formula for alanine is C₃H₇NO₂, meaning its molecular weight would be:

12*3 + 7*1 + 14 + 16*2 = 89 g/mol

Then we divide the sample mass by the molecular weight, to do the conversion:

1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles

4 0

1 year ago

DNA can be replicated through an in vivo mechanism or a polymerase chain reaction (PCR) mechanism. Match each description to the

umka2103 [35]

Complete question from other source attached

Answer:

Explanation:

Catalyzed by DNA polymerase - both. DNA polymerase catalyzes DNA replication in the cell. However, purified versions of the enzyme are also used to synthesise DNA as part of PCR reactions

involves leading strand synthesis only - PCR. In PCR, lagging strand synthesis is not carried out because the DNA is denatured (rendered single stranded). Therefore, each strand is replicated independently by leading strand synthesis

duplicates a small fragment of the genome - PCR. Usually, to carry out PCR, small sequences called primers are used that specify the region of DNA to be replicated

duplicates the entire genome - in vivo replication - when the DNA is replicated in vivo, the entire genome is replicated. This is carried out prior to cell division so that two daughter cells can each inherit a copy of the entire genome

7 0

2 years ago