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Kipish [7]
3 years ago
15

Can someone help me? I will give BRAINLIEST

Chemistry
2 answers:
ELEN [110]3 years ago
6 0
I don’t know the answer but I need you help with my points
Mkey [24]3 years ago
4 0

Troposphere is the highest layer of the atmosphere and touches space

Startophere is where stratus clouds form up high in the sky

The Mesosphere is the third layer you need to know and it’s low density ...yo?

Thermosphere we’re getting closer and closer to earth

Exoshpere were airplanes are high, buckle up it’s time to fly

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5pts. Chemistry multiple choice​
aleksandrvk [35]

Answer:

C. Xenon

Explanation:

The question only requires you to use your visual skills to identify the unknown gas. Look at the spectra of unknown gas and xenon, are they not identical? SInce they are same, the unknown sample must be xenon.

4 0
2 years ago
Read 2 more answers
BRAINLIEST PLEASE HELP!!!
Natalka [10]
Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.
5 0
2 years ago
A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
3 years ago
The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r
spin [16.1K]

<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aniline:</u>

Given mass of aniline = 4.50\times 10^0=4.50g      (We know that:  10^0=1 )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol

  • <u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of acetic anhydride = (1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol

The chemical equation for the reaction of aniline and acetic anhydride follows:

C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = \frac{1}{1}\times 0.048=0.048mol of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = \frac{1}{1}\times 0.048=0.048mol of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0
3 years ago
At which part of a river would you have the best luck growing food?
krek1111 [17]

Answer:

B. Delta

Explanation:

3 0
2 years ago
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