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Andrej [43]
2 years ago
5

You have an unknown compound of Potassium derivatives (KIxOy). This sample contains 17.0g of potassium, and 27.8g of oxygen and

55.2g of iodine. If you prepared another sample, but this time it had 460.0g of potassium periodate, how much oxygen and iodine would you have in your sample?
Chemistry
1 answer:
Alex777 [14]2 years ago
7 0

Answer:

The body temperatures in degrees Fahrenheit of a sample of adults in onle small town are 96.6 99.6 96. 3 97.3 99.8 97.7 97,6 97.9 99.2 Assume body temperatures of adults are normally distributed Based on this data find the 9900 confidence Interval of the mean body temperature of adults in the town Enter Your answer as an Open-interval (i.e. parentheses) accurate to 3 decimal places. Assume the data is from normally distributed population_ 9900 CI = Previewv Tp Enter 4euleie er using intewval notation. Example: [2,5) Uge U fcr unicr to combine interval:, Example: (-00,2] U [4,00} Entet each Faluce nunber (like 33,2.2172) Or 33 calculztion (like 5/3,243 5-4) Enter DNE for ar empty set, Use 00 to enter Infinity. Enter each valze accurate to decial place: Get Help: Points possible: 10 This attempt of 3 Me ye ingtmctor absutthis questin Poelis quesivto 6 EpIC e 6 DeA

Explanation:

(Unsure if this helped but hope it did!)

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An ion has a charge of +1 and 15 protons how many electrons does it have?
VashaNatasha [74]
9 electrons :) I can't give a sbs explanation but here we go
5 0
3 years ago
Read 2 more answers
Organic foods do not contain chemicals.<br><br> True<br> Or<br> False
Mashutka [201]

Answer:

True

Explanation:

The word Organic refers to the methods used to cultivate and process farm agricultural products. Organic foods are edible and nutritious substances consumed (both plants and animals) that are free from the use of synthetics and chemicals. In plants, the include the use of organic manure that serves as fertilizers and carrying out the weeding process by hand weeding. In animals, diseases can be prevented by maintaining a clean house or rotational grazing.

The benefit of organic foods are to produce food substances with no chemical substances.

6 0
3 years ago
Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r
Fudgin [204]

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

7 0
3 years ago
A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca
Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage =  77.77%

Explanation:

<u>Step 1</u>: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

<u>Step 2</u>: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

<u>Step 3</u>: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = <u>0.009725 moles of H2C2O4</u>

<u />

<u>Step 4:</u> Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

<u>Step 5</u>: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

<u>Step 6</u>: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0
3 years ago
The National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. The units the NWS
pashok25 [27]

Answer:

d. 103.3

Explanation:

In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:

In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.

Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.

4 0
3 years ago
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