HELP MEEEEEEEEEEEEEEEEEEEEEEEE

What do all fossil fuels have in common?

Bess [88]

Answer:

Your answer would be D, they are all non-renewable energy sources. Hope this helps!

6 0

3 years ago

The concentration of an additive in a standard sample of palm oil was measured 6 times and the following results were obtained:

Lerok [7]

In the data, 0.20 ppm is an outlier and this can be rejected if there is a 95% confidence level.

<h3>What is an outlier?</h3>

When analyzing data an outlier is a value that is abnormal or too different from other data. In the case presented 0.20 can be tagged as an outlier because other values such as 0.11, 0.12, 0.13, and 0.14 are similar while 0.20 is outside this range.

<h3>Should this piece of data be rejected?</h3>

The general rule is that if there is a 95% of confidence or higher you can reject an outlier, knowing the other data occurs 95% of the time, and therefore the outlier is improbable.

Based on this, you can reject an outlier if the confidence level is 95%.

Learn more about outlier in: brainly.com/question/9933184

7 0

3 years ago

4. A sample of nitrogen at 20 degrees Celsius was compressed to 300ml to

cricket20 [7]

Answer:

We are given:

V(i) = 0.3 L V(f) = 0.36 L

P(i) = x pa P(f) = 400 pa

T(i) = T(f) = 293 k

Using the gas formula:

PV = nRT

Since n , R and T are constant,

PV = k (where k is a constant)

Hence we can say that:

P(i) * V(i) = P(f) * V(f)

x * 0.3 = 0.36 * 400

x = 14.4/0.3

x = 480 pascals

Hence the initial pressure is 480 pascals

Converting to Kpa, we have to divide by 1000

Initial pressure (Kpa) = 0.48

3 0

4 years ago

Please explain how to find each answer solution.

svp [43]

Explanation:

1) Based on the octet rule, iodine form an I⁻ ion.

Therefore,

Option E is correct ✔

2) The electronic configuration of the sulfide ion (S²⁻) is :

₁₆S = 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴

₁₈S²⁻ = 1s² 2s² 2p⁶ 3s² 3p⁶ or [Ne] 3s² 3p⁶

Therefore,

Option E is correct ✔

3) valence shell electron of

Halogens = 7

Alkali metal = 1

Alkaline earth metal = 2

Therefore,

Option D is correct ✔

4) Group 2 element lose two electron in order to achieve Noble gas configuration.

And here Group 2 element is Sr

Therefore,

Option B is correct ✔

5) Group 13 element lose three electron in order to achieve Noble gas configuration.

And here Group 13 element is Al

Therefore,

Option B is correct ✔

6) For a given arrangements of ions, the lattice energy increases as ionic radius decreases and as ionic charge increases.

Therefore,

Option A is correct ✔

7 0

3 years ago

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +

Dominik [7]

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$ $n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$ $V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

.

6 0

2 years ago