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love history [14]
2 years ago
8

Please help 15 points

Chemistry
1 answer:
sleet_krkn [62]2 years ago
7 0
<h3>Further explanation </h3>

Given

Atomic symbol

<h3> Required </h3>

Atomic composition

Solution

Atomic number = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Number of protons and Number of Neutrons  in nucleus

Number of electrons in the shell

From the picture:

protons = 3

neutrons = 4

electrons = 3

atomic number = protons = electrons = 3

mass number = protons + neutrons = 3+4 = 7

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What is the molality of a solution of 10 g NaOH in 500 g water?<br>Molar mass NaOH = 40 g​
umka2103 [35]

Explanation:

Moles of NaOH = 10g / (40g/mol) = 0.25mol.

0.25mol / 500g = 0.50mol / 1000g = 0.50mol/dm³.

The molarity is 0.50mol/dm³.

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3 years ago
If alcohol is ingested rapidly, the _____ will rise sharply within twenty minutes. blood pressure ABC LD50 BAC
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Answer:

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Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?
katovenus [111]

Answer:

Approximately 3.81\; \rm m.

Explanation:

Look up the density \rho of carbon tetrachloride, \rm CCl_4, and glycerol:

  • Density of carbon tetrachloride: approximately 1.59\times 10^{3}\; \rm kg \cdot m^{-3}.
  • Density of glycerol: approximately 1.26\times 10^{3}\; \rm kg \cdot m^{-3}.

Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

P = \rho\cdot g \cdot h.

The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

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3 years ago
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