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3 years ago

How many nonmetals are there in the boron group?

Chemistry

1 answer:

OLga [1]3 years ago

5 0

The answer is going to be there isn't any nonmetals. hope that helped

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A 50.5 L flask contains 3.25 mol of an unknown gas at 288.6 K, what is the pressure of this gas?

Ivahew [28]

Answer:

1.52atm is the pressure of the gas

Explanation:

To solve this question we must use the general gas law:

PV = nRT

Where P is pressure in atm = Our incognite

V is volume = 50.5L

n are moles of gas = 3.25moles

R is gas constat = 0.082atmL/molK

And T is absolute temperature = 288.6K

To solve pressure:

P = nRT / V

P = 3.25mol*0.082atmL/molK*288.6K / 50.5L

P = 1.52atm is the pressure of the gas

3 0

3 years ago

Question 1 of 15 2 Points What is another name for a pure substance? Answer here

maksim [4K]

Answer:

element

Explanation:

elements consist of pure substances

4 0

3 years ago

Can you balance 12 blocks on the 3x2 platform

Jobisdone [24]

Yes, it is possible to go do because it would be 2 stacks of 6

3 0

2 years ago

Which atomic model proposed that electrons move in specific orbits the nucleus of an atom?

MrRa [10]

Bohr's atomic model proposed that electrons move in specific orbits around the nucleus of an atom.

3 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