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3 years ago

Question 1 of 10

Chemistry

1 answer:

Natalka [10]3 years ago

5 0

Answer:

Explanation:

A- incorrect, not in alphabetical order

B- incorrect, symbol for salt in NA not SA

C- correct

D- incorrect, not based on discovery

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6.00g of gold was heated from 20.0c to 22.0c. How much heat was applied to the gold

Alinara [238K]

6 x .129 x 2= 1.55J
q = mass x specific heat x delta T.

5 0

2 years ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

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

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

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

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3 years ago

What is the same for all of the drilling sites we examined?

yarga [219]

Answer:

the same is what is this question like what did u exame

4 0

3 years ago

Read 2 more answers

The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only produ

ohaa [14]

A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67

Then we multiply a number in order to obtain a whole number ratio between the atoms.

1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula

8 0

3 years ago

Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?

RSB [31]

One reason could be that the water in this experiment, didn’t boil, but instead became a “chemical reaction” from atmospheric pressure which is normal.

While heating two different samples, of
sea level water it decides on a temperature of 102°C and the other boils at 99.2°C. Basically calculating of the percent error.

6 0

3 years ago