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AveGali [126]
3 years ago
9

How many moles are in 564 grams of copper

Chemistry
1 answer:
Paraphin [41]3 years ago
7 0
No. of moles =mass given/mass acquired 

no. of moles =564÷64=8.81

therefore no. of moles =8.81m
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The gasses in a hair spray can are at temperature 300k and a pressure of 30 atm, it
Sergeeva-Olga [200]

Answer:

900 K

Explanation:

Recall the ideal gas law:

\displaystyle PV = nRT

Because only pressure and temperature is changing, we can rearrange the equation as follows:
\displaystyle \frac{P}{T} = \frac{nR}{V}

The right-hand side stays constant. Therefore:

\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}

The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.

Substitute and solve for <em>T</em>₂:

\displaystyle \begin{aligned} \frac{(30\text{ atm})}{(300\text{ K})} & = \frac{(90\text{ atm})}{T_2} \\ \\ T_2 & = 900\text{ K}\end{aligned}

Hence, the temperature must be reach 900 K.

7 0
2 years ago
12. If a solution has a pH of 3, what would the pOH value of that solution be? *
butalik [34]

Answer:

The correct answer is pOH= 11

Explanation:

From the aqueous acid-base equilibrium we know that

pH + pOH = 14

If we know pH, we can calculate pOH as follows:

pOH = 14 - pH

In this problem, the solution has a pH of 3, so:

pOH = 14 - 3 = 11

8 0
2 years ago
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What can longitudinal waves travle though?
Sever21 [200]
They travel through all medium types
5 0
3 years ago
23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
2 years ago
1. Consider the decomposition reaction of sodium chlorate. There are 100 grams of
IRINA_888 [86]
A. NaCl(s) and O2(g)

B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)

C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3

D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)

E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl

F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2

G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2

H. Percent yield = 10/45.1 • 100% = 22.2% yield
6 0
3 years ago
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