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
Likurg_2 [28]
3 years ago
10

The substance in a chemical reaction that ate combined or separated to form new substance are the

Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

product

Explanation:

Chemical reactions occur when chemical bonds between atoms are formed or broken. The substances that go into a chemical reaction are called the reactants, and the substances produced at the end of the reaction are known as the products.

i hope it helps please mark me as brainliest :)

You might be interested in
♡
Oksi-84 [34.3K]

Your question has been heard loud and clear.

Both scientific laws and scientific theories are not based on hypothesis.

Because , scientific laws are proven and so they are real not hypothetical.

Whereas scientifc theories can be hypothetical.

Thank you

4 0
3 years ago
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
​If a bag of peas weighs 454 grams, how many peas would be in the bag?
OverLord2011 [107]

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.1 g} = 4540pea

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

454 g \times \frac{1pea}{0.36 g} \approx  1261pea

​In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

You can learn more about conversion factors here: brainly.com/question/1844638

6 0
3 years ago
A solution is prepared by dissolving 0.23 mol of benzoic acid and 0.27 mol of sodium benzoate in water sufficient to yield 1.00
Monica [59]

Answer:

The pH does not decrease drastically because the NaOH reacts with the <u>D) Benzoic acid</u> present in the buffer solution.

Explanation:

The hydroxide ions will react with acidic part of the solutions, it means the benzoic acid, so it will form the conjugate base, the benzoate ion.

7 0
3 years ago
The type of radiation which is identical to a high energy electron is known as a(n). beta, alpha, or gamma particles
Schach [20]
The type of radiation which is identical to a high energy electron is known as a(n) beta.

Answer: A) or the first option.
5 0
3 years ago
Read 2 more answers
Other questions:
  • Is it possible for the overall charge of an ionic compound to be -1.<br> true<br> false
    14·1 answer
  • Your dad is working on creating a brick border for the lake in your backyard. Each brick has a mass of 100 g and a volume of 20
    10·1 answer
  • Calculate the Molecular or Formula massof each of the following:
    8·1 answer
  • Which of the following is not one of the main five types of reactions?
    15·1 answer
  • Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) +
    9·1 answer
  • Complete the sentence using the correct term.
    6·2 answers
  • when water reaches its boiling point and turns into water vapor what happens to its molecular structure
    5·1 answer
  • Dos moléculas de Clorato de Potasio en estado sólido, al aplicarle calor se descompone en dos moléculas de Cloruro de Potasio en
    15·1 answer
  • You buy butter $3.31 a pound one portion of onion compote requires 1.7 oz of butter how much does the butter for one portion cos
    8·1 answer
  • (b) Do compounds of hydrogen exhibit a relatively large or small kinetic isotope effect? Explain.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!