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3 years ago

Round 10235.0 to four significant figuers

1 answer:

8 0

Greetings!

Significant Digits are:
-Non Zero Numbers
-Zeros in between Non Zero Numbers
-Trailing zeros to the right of the decimal point

So when rounding to 4 significant figures, we get:
1023

Hope this helps.
-Benjamin

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A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the iden

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Answer:

HCl

Explanation:

Choices:

CO: 28.01g/mol

NO₂: 46g/mol

CH₄: 16.04g/mol

HCl: 36.4g/mol

CO₂: 44.01g/mol

It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:

PV = nRT

Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).

Replacing:

PV = nRT

PV / RT = n

0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n

8.786x10⁻³ = moles of the gas

As the mass of the gas is 0.320g; its molar mass is:

0.320g / 8.786x10⁻³moles = 36.4 g/mol

Based in the group of answer choices, the identity of the gas is:

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3 years ago

What type of large molecule is responsible for repair and growth of cells?

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Answer:

The answer is a. proteins

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Many metals react with acids to produce hydrogen gas. A certain reaction produces 32.6 L of hydrogen gas at STP. How many moles

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Answer: 1.46moles

Explanation:

Applying PV= nRT

P= 1atm, V= 32.6L, R= 0.082, T = 273K

Substitute and simplify

1×32.6/(0.082×273)=n

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3 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

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Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

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differences:
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chemical transforms rocks into sediments while physical only breaks it down

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(a brainliest would be appreciated)

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