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victus00 [196]
3 years ago
14

Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0.250 kilograms, what should be t

he tension of the string so that it can sustain the weight of the wall clock?

Chemistry
2 answers:
LuckyWell [14K]3 years ago
4 0
T=mg+ma the formula for tension when object not moving.
T=tension
m=mass
g=acceleration of gravity
a=acceleration

a=0 since its not moving.

maks197457 [2]3 years ago
3 0
The tension should be 0.25
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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
Determine the empirical formula of a compound containing 40. 6 grams of carbon, 5. 1 grams of hydrogen, and 54. 2 grams of oxyge
zavuch27 [327]

The empirical formula is C₂H₃O₂

<h3>What is Empirical formula of a compound ?</h3>

The empirical formula is the simplest whole number ratio of elements present in a compound.

The total molar mass of the compound is 118.084 g/mol.

mass of Carbon present = 40.6

mass of Hydrogen present = 5.1 grams

mass of Oxygen present = 2 grams

Moles of C = 40.6/12 = 3.38

Moles of H = 5.1/1.008 = 5

Moles of Oxygen = 54.2/15.999 = 3.38

Ratio of Moles of C to Oxygen is 1 : 1

Ratio of Moles of C to H is 1/1.5

Multiplying each mole fraction by 2

The empirical formula is C₂H₃O₂

To know more about Empirical Formula

brainly.com/question/14044066

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5 0
2 years ago
What is the total number of atoms of magnesium and phosphorus in 3Mg3(PO4)2?
JulsSmile [24]
Since the given formula is . According to cross method formula, magnesium has +2 charge so, is multiplied by 2.
Thus, 1 molecule of magnesium phosphate will contain 2 atoms of phosphorus.
Therefore, three molecules of magnesium phosphate contains following number of atoms.
Mg = 9
P = 6
O = 24
Hence, we can conclude that there are 6 atoms of phosphorus in three molecules of magnesium phosphate, .
4 0
3 years ago
Two unknown household items are being tested with litmus paper. Substance A turns red litmus paper blue and Substance B turns bl
irina1246 [14]

Answer:

Substance A is a base and Substance B is an acid.

According to the experimental results

7 0
3 years ago
A 1.850 g mixture of SrCO3 and SrO is heated. The SrCO3 decomposes to SrO and CO2. What was the mass percentage of SrCO3 in the
Oliga [24]
<span>decomposition of SrCO3 to SrO and CO2 =change in mass

 moles of CO2 =(1.850 g - 1.445 g). 
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
5 0
3 years ago
Read 2 more answers
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