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3 years ago

What is the frame of reference?

Chemistry

1 answer:

erma4kov [3.2K]3 years ago

5 0

Answer:

<em>Frame of Reference</em>

The fixed background surrounding an object, assumed to be at rest and in a fixed position.

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Please help me its balancing the equation

Vika [28.1K]

Answer:

1, 2, 2, 1

Explanation:

6 0

3 years ago

How to write equations and show units for the Total joules gained by water for the Pringle and the Total Joules per gram of food

Paraphin [41]

Answers:

1. 7500 J; 3800 J/g

Step-by-step explanation:

1. Joules gained by water

q = mcΔT

Data:

m = 100 g

C = 4.184 J·°C⁻¹g⁻¹

ΔT = 18 °C

Calculation:

q = 100 × 4.184 × 18 =7500 J

2. Joules per gram of Pringle

Energy gained by water = energy lost by Pringle

q = -mΔH

7500 J = -1.984 g × ΔH

ΔH = -7500 J/1.984 g = -3800 J/g

Pringles contain 3800 J of food energy per gram.

7 0

3 years ago

What causes the shielding effect to remain constant across a period?

Ivanshal [37]

Electrons are added to the same principal energy level.

5 0

4 years ago

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

The theoretical yield of zinc oxide in a reaction is 486 g. What is the percent

PtichkaEL [24]

Answer:

the correct answer is c

Explanation:

becuase i had the same question

8 0

3 years ago