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Ostrovityanka [42]

3 years ago

8

An element has six valence electrons available for bonding.which group of the periodic table does this element most likely belon

g to

2 answers:

sesenic [268]3 years ago

6 0

It would be group 16

mihalych1998 [28]3 years ago

4 0

Group 16 is the correct answer!

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Can someone help me pls

ella [17]

Answer:

B or C

Explanation:

the two maybe try it

4 0

3 years ago

During an ice age, land can move and specific areas can be permanently changed. true or false

givi [52]

Answer: The given statement is true.

Explanation:

It is known that due to increase in temperature ice melts. Therefore, during ice age there is also melting of ice and solid state of water changes into liquid state of water.

Therefore, this water moves from its initial place and changes its position result in the change of land and specific areas.

Thus, we can conclude that the statement during an ice age, land can move and specific areas can be permanently changed is true.

5 0

3 years ago

Which of these does not describe the Wright brothers' first attempts at flight?

Vlad [161]

They mimicked the flight of birds.

3 0

2 years ago

The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca

geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

$M=d\times V=1.00 g/mL\times 150 ml =150.0 g$

Mass of solution = m = M = 150.0 g

$NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol$

Moles of NaOH = $\frac{5.00 g}{40 g/mol}=0.125 mol$

Energy released when 0.125 moles of NaOH added in water = Q

$Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J$

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

$Q'= 5,563.8 J$

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = $\Delta T$

$Q'=mc\times \Delta T$

$5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T$

$\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC$

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0

3 years ago

Mn(OH)2(s) + MnO4(aq) → MnO42–(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what

PilotLPTM [1.2K]

Answer:

Hi, the given equation has some missing parts. Actual equation is- ' $Mn(OH)_{2}(s)+MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$ '

balanced equation: $Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Explanation:

$Mn(OH)_{2}(s)\rightarrow MnO_{4}^{2-}(aq.)$

Balance O and H in basic medium: $Mn(OH)_{2}(s)+6OH^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Balance charge: $Mn(OH)_{2}(s)+6OH^{-}(aq.)-4e^{-}\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$ ........(1)

$MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$

Balance charge: $MnO_{4}^{-}(aq.)+e^{-}\rightarrow MnO_{4}^{2-}(aq.)$ .....(2)

$[equation(2)\times 4]+[equation (1)]:$

$Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

$OH^{-}(aq.)$ is present on the left hand side of balanced equation and it's coefficient is 6

6 0

2 years ago