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3 years ago

How many atoms are in 5Cao4?

Chemistry

1 answer:

vladimir1956 [14]3 years ago

5 0

Answer:

There are 30 atoms in total

Explanation:

There is 5 carbon, 5 of whatever A is and 4 oxygens but since the whole symbol has been multiplied by 5 there would be 20 oxygen

So 20 + 5 + 5 = 30 atoms

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which of these sentences is true about nitrogen? a- as nitrogen is the rarest gas in the atmosphere, very few organism can use i

Nataly [62]

B is true because plants can only use nitrogen in the form of nitrates, no other living thing needs nitrogen therefore the percentage of nitrogen in the atmosphere is 79%

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3 years ago

Two samples of gold that have different temperatures are placed in contact with one another. Heat will flow spontaneously from a

galina1969 [7]

It is 50. This is because heat goes from higher temperatures to lower and nothing else in the options is lower.

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3 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

Which example is not a structural adaptation of a plant? A. curling around a tree trunk B. protecting itself with thick bark C.

aalyn [17]

A. curling around a tree trunk

I believe would be the correct answer because it is not an adaptation that it has gained through evolution and it is something that happens after the plant has grown.

3 0

3 years ago

1. 50.0 g of an unknown metal was heated to 95.0 °C and added to 45.0 g of water at 25.0 °C

Soloha48 [4]

Answer

Q = mcΔT

Qwater = -Qmetal

(100 g)(4.18 J/g°C)(25°C - 20°C) = -(50 g)c(25°C - 90.0°C)

2090 J = (3250 g °C)c

c = 0.643 J/g °C

8 0

2 years ago