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3 years ago

Which of the following measurements are recorded to three significant figures? Select all that apply.

Chemistry

2 answers:

irinina [24]3 years ago

7 0

Answer:

c) 1610 s

d) 0.0890 m

Explanation:

Significant figures:

The given measurement have four significant figures 1234.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side when decimal point is present e.g 2400. are also significant. There are four significant figures are present.

a) 0.003 L

This measurement have 1 significant figure which is 3.

b) 140 g

This measurement have 3 significant figure 1, 4. In this measurement 0 is trailing zero that's why it is not consider significant.

c) 1610 s

This measurement have 3 significant figures 1, 6, 1.

d) 0.0890 m

This measurement have 3 significant figures 8,9 and 0

Kay [80]3 years ago

4 0

C and D have three significant figures.

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Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.

marta [7]

Answer:

Approximately $53.3\; \rm g$ .

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$ , $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm H$ : $1.008$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$ :

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$ .

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6 0

3 years ago

A student was provided with a solid sample made up of a mixture of components. Prior to separation, the student measured 2.895 g

Shtirlitz [24]

Answer:

Component 1, since it has the greatest mass.

Explanation:

Hello,

In this case, considering that the by-mass percent of a substance is computed:

$\% m/m_i=\frac{m_i}{m_{mixture}}*100\%$

In such a way, given the form of the equation one could infer that the higher the mass of the i component, the higher its by-mass percent, it means that the component with the greatest mass will have the highest by-mass percent for the mixture. In such a way, since the component 1 was found to have a mass of 1.12 g, it will have the highest by-mass percent.

Best regards.

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