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lara [203]
3 years ago
10

As the atomic number of elements increases across a period on the periodic table, the atomic radius

Chemistry
1 answer:
zheka24 [161]3 years ago
7 0

Answer:

Decreases  

Explanation:

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If you have 8.943 L of carbon monoxide gas at SATP, how many moles would it contain?
viktelen [127]
1mole contains 22.4Lmol^-1
xmole contains 8.943
cross-multiply
x=1×8.943/22.4
x=0.40mole
there it contains 0.40moles.
8 0
3 years ago
¿Cuál de las siguientes configuraciones globales corresponde a un elemento químico que se comporta como metal? 1 punto [He]2s2 2
yuradex [85]

Answer:

[Ne]3s2

Explanation:

ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.

considerando la configuración electrónica más externa de cada una de las especies mostradas;

para la primera configuración, ns2 np6 corresponde a un gas noble.

para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.

para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.

para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6

5 0
3 years ago
Can someone please list one of the groups of representative elements in 1A-7A
Vadim26 [7]

Group 1A(1), the alkali metals, includes lithium, sodium, and potassium. Group 7A(17) the halogens, includes chlorine, bromine, and iodine. hope this helps:)

7 0
3 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
Consider the balanced equation for the following reaction:
Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g

To calculate the theoretical yield of lithium chlorate, we use the equation:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0
3 years ago
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