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3 years ago

Which element and how many atoms does each compound have in the picture?

Chemistry

1 answer:

Stells [14]3 years ago

7 0

The first one has 35 atoms and the elements hydrogen, oxygen and chlorine

The second one has 9 atoms and the elements sodium, carbon, and helium

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How are electrons arranged around an atom?

Schach [20]

They are arranged in shells

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2 years ago

The morning temperature in a city is 41°F. If a sunny, mild day is forecast, which temperature is most likely for 2:00 p.m.?

romanna [79]

Answer:

huh

Explanation:

5 0

2 years ago

How do you test for carbon dioxide ?? Explain

g100num [7]

Answer:

Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate

Explanation:

. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white

5 0

2 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{100}{1000}kW=0.1 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.1 kW\times 2190 h=219 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

$Cost=\$ (219\times 0.15)=\$ 32.85$

Therefore, annual cost of incandescent light bulb is $\$ 32.85$

(b) Power of bulb is 25 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{25}{1000}kW=0.025 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.025 kW\times 2190 h=54.75 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

$Cost=\$ (54.75\times 0.15)=\$ 8.21$

Therefore, annual cost of fluorescent bulb is $\$ 8.21$ .

7 0

3 years ago