All categories

2 years ago

How has the U.S. Census Bureau changed the way it records information about relationships?

1 answer:

8 0

U.S. Census Bureau changed the way it records information about relationships by increasing the number of relationship categories to include same-sex partner options and is denoted as option B.

<h3>What is Relationship?</h3>

This is a close connection between people and is characterized by physical or emotional intimacy.

There has been changes in the way U.S. Census Bureau records information is recorded in this aspect by increasing the number of relationship categories for inclusiveness of other types of people such as those who have same-sex partners and so on.

This is therefore the reason why option B was chosen as the correct choice.

Read more about Relationship here brainly.com/question/10286547

#SPJ1

You might be interested in

Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

Elena-2011 [213]

Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

$\rm Ca$ : $40.078$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

$\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}$ .

Notice that in the balanced chemical reaction, the ratio between the coefficient of $\rm HCl$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}$ .

In other words, each mole of $\rm HCl$ molecules consumed would produce only $0.5\;\rm mol$ of $\rm CO_2$ molecules.

Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

If $\rm CaCO_3$ is the limiting reactant, $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. However, if $\rm HCl$ is the limiting reactant, $1.25\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be produced.

In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .