HELP! HELP! an element with 5 protons, and 8 electrons has an atomic number of?

The metric unit for volume is the? A. pound B.quart C.liter D.gallon

Vladimir [108]

I believe it is liter.

5 0

3 years ago

Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas? A)He(g)

umka21 [38]

Answer: He(g)

Explanation: I had the same question and I got the answer right

8 0

3 years ago

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

3 years ago

What is most likely the amount of energy available at a trophic level of tertiary consumers if the amount of energy available to

meriva

Answer:

D.) 2,000 kilocalories would be the correct answer.

Explanation:

4 0

2 years ago

Read 2 more answers

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

but in case of NaOH and HCl n-factor is 1 for each.

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

7 0

3 years ago