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icang [17]
3 years ago
11

For which of the following materials is necessary to stop a beta particle?

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
Hello.

The answer would be <span>A. Three feet of concrete.

Have a nice day.</span>
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The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
4 0
3 years ago
Please help me it is easy I will reward Brainliest
Dvinal [7]

Answer:

Parallel Circuit

8 0
3 years ago
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How are simple cations and anions named ?
algol13
Simple cations are formally called by their element names with a suffixed Roman numeral in parentheses to indicate its charge. A simple anion has a name that is the original elemental name with the final syllable changed to -ide.
3 0
3 years ago
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The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o
telo118 [61]
If you clear volume in the density equation:

\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3

V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3
6 0
3 years ago
How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the wor
tangare [24]
Answer is: 4.45 grams of methane gas <span>need to be combusted</span>.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V = n·R·T.<span>
p = 1.1 atm.
T = 301 K.
V(H</span>₂O) <span>= 12.5 L.
R = 0,08206 L·atm/mol·K.
</span>n(H₂O) = <span>1.1 atm · 12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
</span>n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.
4 0
3 years ago
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