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Sophie [7]
3 years ago
8

Express (158 lb)2 to three significant figures with appropriate units

Chemistry
2 answers:
statuscvo [17]3 years ago
5 0
(158lb)2 = 158 lb * 158 lb 
This gives us 24964 lb^2 
To solve the solution to 3sf we obtain 25000 lb^2. 
Hence 25000 to 3sf
Setler [38]3 years ago
5 0

Answer: 2.50 × 10⁴


Justification:


1) The expression is: (158 lbs)²


2) When two amounts are multiplied, the product must be shown with the same number of significant figures of that of the factor with the least number of significant figures.


3) So you have to muliply 158 by itself (square) and then round to 3 significant figures.


4) The number 158 has 3 significan figures, so the product shall show also 3 significant figures:


(158 lb)² = 158 lb × 158 lb = 24,964 lb², which must be rounded to 3 significant figures: 25,000.


Since, the fourth figure was greater than 5, the third signficant figure must be increased. Since the third signficant figure is 9, the increase is translated to the second significant figure.


5) Now, to indicate tha the number of significan figures is 3, you have to use scientific notation:


25,000 = 2.50 × 10³, where the 3 signficant figures are 2, 5, and 0.

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6 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
Is the cyanide ion a strong or weak base?
11111nata11111 [884]
It's a weak base bacause H C N is weak
4 0
3 years ago
Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and Tr
ale4655 [162]

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

   $=\frac{x^2}{0.02 -x}$

  $=8.32 \times 10^{-9}$

Clearing $x$, we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$, one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

            $= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.

3 0
3 years ago
Why is wood both renewable and nonrenewable?​
Tasya [4]

Answer:

"Wood is technically a renewable source of energy, since it is a form of biomass. All biomass is traditionally considered a renewable source of energy because it is part of the natural carbon cycle on Earth." I found this from https://www.quora.com/

Explanation:

4 0
3 years ago
Read 2 more answers
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