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pochemuha
3 years ago
7

How many grams is in a 5 pound sample? express your answer in scientific notation with four significant figures. (1 pound = 453.

592 grams)?
A) 2.3 x 10^3 g
B) 1.1 x 10^4 g
C) 2.27 x 10^2 g
D) 2.268 x 10^3 g
Chemistry
2 answers:
marishachu [46]3 years ago
8 0

Answer:

D) 2.3 x 10^3 g

Explanation:

to know how many grams are in 5 pound its necesary to multiply the weight by the unit conversion factor 453.592 grams/pound

5 pound * 453.592 grams / pound = 2267.96 grams

because it is necesary to express the answer in scientific notation with four significant figures, the value obtained initially is expressed in scientific notation (1)  and then rounded up with these significant figures (2)

2267.96 grams = 2.26796 x 10^3 (1)

2.26796 x 10^3 ----> 2.268 x 10^3 (2)

d1i1m1o1n [39]3 years ago
3 0
The answer is A. I hope this helps
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C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were
melomori [17]

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂  +  2HCl     →     PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number  of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

                            PbCl₂        :          Pb(NO₃)₂

                               1             :             1

                             0.23         :            0.23

Mass of Pb(NO₃)₂:

Mass = number of moles ×  molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

8 0
4 years ago
A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th
Ann [662]

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

            Q = M x C x ΔT

            Q = -100 x 4.18 x (-5.14)

            Q = 2148.5 joules

            Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

          ΔH = Q / number of moles of NaHCO3

          ΔH = 2.1485 / 0.173

          ΔH = 12.42 KJ/mol

3 0
3 years ago
Convert 2.072x10^43 atoms element to moles
tangare [24]

Given :

Number of atoms of an element, n = 2.072 × 10⁴³ atoms.

To Find :

Number of moles of that element in given number of atoms.

Solution :

We know, 1 mole of any element contains 6.022 × 10²³ atoms.

So, number of moles in given number of atoms are :

n = \dfrac{2.072\times 10^{43}}{6.022\times 10^{23}}\\\\n = 3.441 \times 10^{19}\ moles

Hence, this is the required solution.

8 0
3 years ago
In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
4 years ago
List the three principal states of matter, give two examples of each
icang [17]

matter can take be one of the three states which are; (Solid, Liquid, Gas)

Think about water, it can be a Solid (Ice) a Liquid (not frozen obviously) and a Gas (when it evaporates)

7 0
3 years ago
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