When baking soda and vinegar react to certain surfaces up bubbles what does this mostly indicate

25 points! Help please

Sedbober [7]

Answer:

I think its b

Explanation:

but I wouldn't depend on this answer

4 0

3 years ago

Read 2 more answers

Which sentence is a scientific statement?

LekaFEV [45]

Solution: Sentence D is a scientific statement.

Scientific statement is a statement that defines the science behind any situation. There should be any experimental result or observation to state any scientific statement. Also, the statement should be accepted by considering other scientific aspects.

Here, statement A can not be a scientific statement as different aromas of cooked food are liked by different people. Any one can find aroma of cooked food in copper better than the ceramic pot.

Thus, the statement can be proved wrong easily.

Statement B can not be a scientific statement as people can find the taste of normal water (without ice cubes) better than the ice cold water.

Statement C is not a scientific statement as there is no science behind watching baseball or ice hockey, it depends on the liking of today's people.

Statement D is a scientific statement because this can be scientifically proved.

5 0

4 years ago

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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5

Citrus2011 [14]

Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}years^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = 100

a - x = amount left after decay process = $\frac{70}{100}\times 100=70$

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}$

$t=2948years$

Thus the fossil is 2948 years old.

5 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

(a) Show that the pressure exerted by a fluid P (in pascals) is given by P = hdg, where h is the column of the fluid in metres,

Komok [63]

Answer:

B) THE DEPTH OF THE LAKE IS 0.060 m

Explanation:

b) Determine the depth of the lake in metres

1. Using the general gas law, we will calculate the initial pressure of the air bubbles.

P1V1 /T1 = P2V2/T2

P1 = Unknown

T1 = 5.24 °C

T2 = 18.73 °C

P2 = 0.973 atm

V1 = V1

V2 = 6V1

P1 = P2 V2 T1 / V1 T2

P1 = 0.973 * 6V1 * 5.24 / V1 * 18.73

P1 = 5.09852 * 6 / 18.73

P1 = 30.59112 / 18.73

P1 = 1.633 atm.

2. Calculate the depth of the lake:

Pressure = length * density * acceleration

length = Pressure / density * acceleration

Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2

Density = 1.02 g/cm3 = 1.02 * 10^3 kg/m^3

Acceleration = 9.8 m/s^2

So therefore, the length in metres is:

Length = density * acceleration / pressure

Length = 1.02 *10^3 * 9.8 / 165, 463.725

Length = 9.996 * 10^3 / 165 463.725

Length = 0.06 m

Hence, the depth of the lake is 0.06 m

6 0

3 years ago