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3 years ago

State the number of sig figs in each value:

Chemistry

1 answer:

Rasek [7]3 years ago

8 0

Answer:

Explanation:

Significant figure implies number of digits that are to be considered. Some rules are required to be considered when writing a given expression to an expected significant figures.

So that:

1) 0.00004050 is 4 significant figures

2) 54.7000 is 6 significant figures

3) 1,000.09 is 6 significant figures

4) 0.039 is 2 significant figures

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Solid iron is mixed with a solution of copper (I) nitrate to form iron (III) nitrate solution and metal copper. (what is the equ

DedPeter [7]

Answer:

Fe + 3CuNO₃ → Fe(NO₃)₃ + 3Cu

Explanation:

Solid Iron = Fe

Copper (I) nitrate = CuNO₃ (Nitrate, NO₃⁻, always has a charge of -1).

Iron (III) nitrate = Fe(NO₃)₃ (That way the compound has an overall neutral charge)

Metal Copper = Cu

Writing the equation using symbols leaves us with:

Fe + CuNO₃ → Fe(NO₃)₃ + Cu

It is not balanced yet. Now we balance the NO₃ species on the left side:

Fe + 3CuNO₃ → Fe(NO₃)₃ + Cu

Finally we balance the Cu species on the right side:

Fe + 3CuNO₃ → Fe(NO₃)₃ + 3Cu

5 0

2 years ago

Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r

Fudgin [204]

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

7 0

3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

Consider the following intermediate chemical equations.

vfiekz [6]

Answer:

-250.3kJ

Explanation:

Based in the reactions and using -Hess's law-:

(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ

(2) 4PCl₅(g) → P₄(s) + 10Cl₂ ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ

Dividing this reaction in 4:

PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = -250.3kJ

8 0

3 years ago

Why are significant figures rules for calculations more important in science than in math class?

inysia [295]

Answer:

The importance of significant figures

As stated before, it is important within the science fields that you are not more precise or accurate than the least accurate or precise number. In science, it is generally agreed upon that the last number digit in any figure is filled with uncertainty.

Explanation:

4 0

3 years ago