1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mazyrski [523]
3 years ago
13

If Mary likes Johnny, but Johnny likes Katie, and Katie likes Mary. What do they do?

Chemistry
2 answers:
RUDIKE [14]3 years ago
8 0

Answer:

love triangle?- orr they can figure something out

elena-s [515]3 years ago
4 0

Answer:

Russian Roulette?

You might be interested in
Among the groups of elements listed below, which have the same number of electrons in their outermost energy levels? C, K, Ca, R
tatyana61 [14]

Answer:

i believe its d

Explanation:

3 0
3 years ago
Read 2 more answers
What is a mutation?
Masja [62]

Answer:

the answer is A hope this helps

3 0
3 years ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0
3 years ago
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
4 years ago
Give an example of a change that would be a chemical change but not a physical change.
Annette [7]

Answer:

digesting food

Explanation:

you dont see it

4 0
3 years ago
Read 2 more answers
Other questions:
  • What are three moving parts of electricity
    15·1 answer
  • 3. Based on the Periodic Table, which of the following elements has physical
    10·1 answer
  • If I apply 0.108 kJ of energy in order to increase the temperature of a bar of gold from 30.0°C to 34.7°C, and the specific heat
    15·1 answer
  • What is Mendeleev most known for?
    8·1 answer
  • Find the energy of a photon of light if the wavelength of the light wave is 6.88 x 10–7m
    11·1 answer
  • What is the mass of oxygen gas in a 16.8 l container at 23.0◦c and 2.50 atm? answer in units of g?
    9·1 answer
  • How many kilograms of water at 15oC can be heated to 95oC by burning 143 grams of methane, CH4, assuming that 100% of the heat i
    10·1 answer
  • Jose times how long sugar takes to dissolve in warm water. He conducts four trials of his experiment. What should he conclude fr
    13·1 answer
  • Can someone help me with this question. <br><br> I’ll mark brainliest
    6·1 answer
  • Air pressure is measured in pascals. For a professional American football game, the ball should be inflated to about 90,000 pasc
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!