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Shkiper50 [21]
3 years ago
12

If 10 staples have a mass of 1.33 g, what will be the mass of 225 staples?

Chemistry
2 answers:
kenny6666 [7]3 years ago
7 0

Answer: A

Explanation: 1.33/10 = .133

.133* 225 = 29.925 which is about 29.9 grams

Leona [35]3 years ago
5 0

Answer:

A) 29.9g

Explanation:

first find the weight of 1 staple.

then multiply with 225

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Which the following countries founded their first permanent settlement in North America: England, the Netherlands,France,and Spa
butalik [34]
England with the Jamestown colony
8 0
3 years ago
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
Calculate the mass of xenon difluoride gas with a volume of 0.223 L, pressure of 0.799 atm and temperature of 47.0 oC.
Greeley [361]

The mass of the gases can be determined by the moles of the gas in the ideal equation. The mass of xenon difluoride at 0.799 atm is 0.011 gms.

<h3>What is an ideal gas equation?</h3>

An ideal gas equation gives the moles of the substance from the temperature, volume, and pressure of the gas. The ideal gas equation can be shown as:

n = PV ÷ RT

Here, n = mass ÷ molar mass

Given,

Volume of xenon difluoride (V) = 0.223 L

Pressure of xenon difluoride (P) = 0.799 atm

Temperature of xenon difluoride (T) = 320.15 Kelvin

Gas constant (R) = 8.314 J⋅K⁻¹⋅mol⁻¹

The moles of the gas is calculated as:

n = PV ÷ RT

= 0.223 × 0.799 ÷ 8.314 × 320.15

= 0.1781 ÷ 2661.72

= 6.69 × 10⁻⁵ moles

Mass is calculated as:

169.29 × 6.69 × 10⁻⁵ = 0.011 gm

Therefore, 0.011 gms is the mass of xenon difluoride.

Learn more about ideal gas here:

brainly.com/question/14189621

#SPJ1

8 0
2 years ago
How many electrons fit in the part of the third energy level shown
makkiz [27]

Answer:

18 electrons

Explanation:

Note: The third energy level can actually hold up to 18 electrons, so it is not really filled when it has 8 electrons in it.

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2 years ago
HELP ME
krok68 [10]

Answer:

D group

................

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