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3 years ago

17 p 18 n What is the atomic number of this element?

Chemistry

1 answer:

Verdich [7]3 years ago

6 0

Answer:

Chlorine

Explanation:

The P usually stands for Protons, and that is usually the element number. (chlorine is 17) N is for neutron

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To convert from liters/second to cubic gallons/minute, multiply the number of liters/second by 15.850 0 0.0353 00.2642 0 60

ivanzaharov [21]

Answer: 15.850

Explanation:

The conversion used from liters to gallons is:

1 L = 0.264172 gallon

The conversion used from sec to min is:

60 sec = 1 min

1 sec = $\frac{1}{60}\times 1=0.017min$

We are asked: liters/sec = gallons/min

$liters/sec=\frac{0.264172}{0.017}=15.850gallons/min$

Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.

8 0

3 years ago

Isotopes of the same element have the same chemical properties because they have the same number of

Sloan [31]

Answer:

PROTONS

Explanation:

Isotopes of the same element have the same chemical properties because they have the same number of PROTONS (and electrons).

4 0

3 years ago

Who ever answer this is genius ????????? let's see?

Flura [38]

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3 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

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

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

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

What is the oxidation number of sulfur in H2SO3? a. +3 b. +1 c. +2 d. +4

Alla [95]

We need to keep in mind that the compound is neutral.

H2SO3
2(+1)+S+3(-2)=0 (since its neutral)
2+S-6=0
S-4=0
S=4

Therefore the oxidation number for sulfur is +4.

3 0

3 years ago

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