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3 years ago

Please help

Chemistry

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

$1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L$

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given $BaSO_{4}$ solution is as follows.

$Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g$

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

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(CH3)2-CH-CH2-O(CH3)3IUPAC NAME

bazaltina [42]

Answer:

1-(tert-butoxy)-2-methylpropane

Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

3 0

3 years ago

Help with atoms WILL MARK BRAINLIEST

Eddi Din [679]

Answer:

C 1:1

Explanation:

Hydrogen loses an electron to becone +1 which is a cation and flourine gains that electron to have a full outer shell on the 2p subshell to become -1 and is an anion so the ratio is 1:1.

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2 years ago

Carbon-14 has a half-life of approximately 5700 years. How much of a 1000g sample will be left undecayed and still radioactive a

Artist 52 [7]

Answer:

option C is correct (250 g)

Explanation:

Given data:

Half life of carbon-14 = 5700 years

Total amount of sample = 1000 g

Sample left after 11,400 years = ?

Solution:

First of all we will calculate the number of half lives passes during 11,400 years.

Number of half lives = time elapsed/ half life

Number of half lives = 11,400 years/5700 years

Number of half lives = 2

Now we will calculate the amount left.

At time zero = 1000 g

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At second half life = 500 g/2 = 250 g

Thus, option C is correct.

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3 years ago

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Chemical change? Need more context.

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3 years ago

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Makovka662 [10]

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