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frez [133]
2 years ago
14

A science class did an experiment in which two substances were mixed. After 15 minutes, everything looks the same, but the stude

nts noticed an odor coming from the experiment. What does this likely indicate?
a.) a liquid was formed.
b.) a temperature change occurred
c.) a chemical reaction occurred
D.) a physical change occurred
Chemistry
2 answers:
valentinak56 [21]2 years ago
5 0

Answer:

C a chemical reaction formed

Explanation:

if something changes smell and nothing else it is a chemical reaction, every other answer does not make sense

Setler [38]2 years ago
5 0

Answer:

chemical reaction

Explanation:

nothing changed with the appearance of how it looks

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Identify the parts of the atom that are labeled in the diagram
baherus [9]

Answer:

A- Nucleus

B-Energy levels

Explanation:

  • An atom refers to the smallest particle of an element that can take part in a chemical reaction.
  • An atom is made up of two major parts namely; the nucleus and the energy shell or energy levels.
  • The nucleus of an atom contains the sub atomic particles, protons and neutrons while the energy shells contain the sub atomic particle electrons.
  • It is important to note that the protons in the nucleus are positively charged which makes the nucleus to be positively charged.
  • On the other hand, energy shells are negatively charged because of the negative charge of the electrons.
6 0
3 years ago
Read 2 more answers
Ordinarily, when we talk about green infrastructure, what type of public element are we most likely to be discussing?
Hitman42 [59]
Parks and open spaces
hope that helps if u have any questions let me know and also if u could mark this as brainliest i would really appreciate it!
6 0
2 years ago
Tuliskan persamaan tetapan kesetimbangan untuk reaksi-reaksi berikut a. Fe3+(aq) + SCN– (aq) ↔ FeSCN3+(aq) b. 3Fe(s) + 4H2O(g) ↔
nevsk [136]

Answer:

a. K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Explanation:

Untuk semua jenis reaksi umum:

aA + bB\iff cC + dD

Konstanta kesetimbangan K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}

Dari pertanyaan yang diberikan:

a. Fe3^+_{(aq)} + SCN^-_{ (aq)} \iff FeSCN^{3+}_{(aq) }

Konstanta kesetimbangan:

K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. 3Fe_{(s)} + 4H_2O_{(g)} \iff Fe_3O_4_{(s)} + 4H_{2(g)}

Konstanta kesetimbangan untuk tekanan parsial K_p

K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Karena Fe3O4 (s) hadir sebagai padatan.

8 0
2 years ago
Coke is an impure form of carbon that is often used in the in- dustrial production of metals from their oxides. If a sample of c
steposvetlana [31]

Answer:

The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.

Explanation:

2CuO+C\rightarrow 2Cu+CO_2

1 Ton = 907185 grams

Mass of copper oxide = 1.0 Ton = 907185 grams

Moles of copper oxide =\frac{907185 g}{79.55 g/mol}=11,403.95 moles

According to reaction, 2 moles of copper oxide reacts with 1 mole of carbon.

Then 11403.95 moles of copper oxide will react with:

\frac{1}{2}\times 11403.95 mol=5,701.98 mol of carbon

Mass of 5,701.98 moles of carbon:

5,701.98 mol\times 12 g/mol=68,423.75 g

Mass of coke = x

Mass of carbon = 68,423.75 g

Percentage of carbon in coke = 95%

95\%=\frac{68,423.75 g}{x}\times 100

x=720,250.09 g=0.794 Ton

The mass of coke needed to react completely with 1.0 ton of copper(II) oxide is 0.794 Ton.

3 0
3 years ago
If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0
2 years ago
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