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Thepotemich [5.8K]
3 years ago
9

PLZZZZ HELP!!!!!! WILL GIVE BRAINLIEST!!

Chemistry
2 answers:
guapka [62]3 years ago
4 0
Color change bubbles and i think dissolving
Jlenok [28]3 years ago
4 0

The formation of bubbles and a colour change are signs of a chemical change.

One of the substances must have evolved the gas and changed into something else.

One of the substances must have changed into something else with a different colour.

Both above are chemical changes.

<em>Evaporation</em> and <em>dissolving</em> are <em>physical changes</em>, because the substances do not change their chemical structure in the process.

You might be interested in
A sample of air is slowly passed through aqueous Sodium hydroxide and then over heater copper. Which gases are removed by this p
bagirrra123 [75]

Carbon dioxide and oxygen are removed from the air.

Explanation:

When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.

First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :

CO₂ + H₂O → H₂CO₃

The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):

H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O

After this by passing the air over heated cooper the oxygen is removed.

2 Cu + O₂ → 2 CuO

Learn more about:

neutralization reaction

brainly.com/question/2632201

#learnwithBrainly

7 0
3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
3 years ago
Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO
sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂   → CaCO₃

Number of moles of CaO:

Number of moles  = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles  = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

                  CO₂         :                CaCO₃  

                  1               :                 1

                 0.31           :              0.31

                CaO           :               CaCO₃  

                 1                :                 1

                 0.26         :              0.26

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ =  26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

7 0
3 years ago
PLEASE HELPP!!!!
Alenkinab [10]

Answer:

Na the valency electrons is 1

The ion charge is positive 1

K the valency electrons is 1

The ion charge is positive 1

Explanation:

Mg the valency electron is 2 The ion charge is positive 2

Ca the valency electron is 2 The ion charge is positive 2

Al the valency electron is 3 The ion charge is positive 3

O the valency electron is 2 The ion charge

is negative 2

F the valency electron is 1 The ion charge is negative 1

Cl the valency electron is 1 The ion charge

is negative 1

N the valency electron is 3 The ion charge is negative 3

S the valency electron is 2 The ion charge is

negative 2

3 0
3 years ago
Ill give u brainliest help asap
Anna007 [38]

I think its A. it cant be C or D b/c we are measuring the distance between Earth and Saturn, not the speed.

7 0
2 years ago
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