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3 years ago

Science begins with an observation, true or false?

Chemistry

1 answer:

JulijaS [17]3 years ago

7 0

Answer:

true!

Explanation:

if youre going by the scientific method, observations are mainly first. then questions, research, etc

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Help and show work please

olya-2409 [2.1K]

ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :

•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32

•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius

•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin

4 0

3 years ago

Does graphite dissolve in cyclohexane

Brums [2.3K]

Answer: Graphite is nonpolar carbon which associates with the nonpolar hexanes. The cellulose in the exposed paper contains polar regions that are attracted to the polar water.

Explanation:

3 0

2 years ago

When may a scientific theory be revised?

Anika [276]

Answer:

it can be revised at any time.

Explanation:

6 0

4 years ago

Read 2 more answers

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago

A fat is composed of long chains of carbon and hydrogen atoms. In a reaction with a strong base, a fat forms a soap and glycerol

Nonamiya [84]

Answer:

Empirical formula is C₉H₁₅O

Molecular formula = C₈₁H₁₃₅O ₉

Explanation:

Percentage of carbon = 77.60%

Percentage of oxygen = 11.45%

Percentage of hydrogen = 10.95%

Molecular weight = 1253 g/mol

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of C = 77.60 g /12g/mol =6.5

Number of gram atoms of O = 11.45 g / 16 g/mol = 0.72

Number of gram atoms of H = 10.95 g / 1.008 g/mol= 10.9

Atomic ratio:

C : H : O

6.5/0.72 : 10.9/0.72 : 0.72/0.72

9 : 15 : 1

C : H : O = 9 : 15 : 1

Empirical formula is C₉H₁₅O

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 1253 / 139

n = 9

Molecular formula = n (empirical formula)

Molecular formula = 9 (C₉H₁₅O )

Molecular formula = C₈₁H₁₃₅O ₉

4 0

3 years ago