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aleksandr82 [10.1K]
2 years ago
6

Add 3 x 10^9 +5.3 x 10^10

Chemistry
2 answers:
Ghella [55]2 years ago
8 0
<span>Add 3 x 10^9 +5.3 x 10^10

</span><span>   3 x 10^9
 +5.3 x 10^10
</span>------------------
5.6 x 10^10
Oksana_A [137]2 years ago
8 0

Answer:

3*10⁹ + 5.3*10¹⁰ = 5.6*10¹⁰

Explanation:

The values are given as exponents:

3*10⁹ and 5.3*10¹⁰

As per the rules:

-Before adding two numbers, all exponents must be the same

-The addition takes place between the numerical portion of the numbers

- The final value is rounded off to the least places in the decimal portion of the given numbers

Therefore, the two numbers to be added are:

3*10^{9}+53*10^{9}=56*10^{9}

Rounded off the final value to 1 decimal place gives:

5.6*10¹⁰

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3 years ago
Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume
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Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

             mass = 0.508 g,               Volume = 0.175 L

             Temperature = (25 + 273) K = 298 K,       P = 1 atm

As per the ideal gas law, PV = nRT.

where,  n = no. of moles = \frac{mass}{\text{molar mass}}

Hence, putting all the given values into the ideal gas equation as follows.

               PV = \frac{mass}{\text{molar mass}} \times RT            

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4 0
3 years ago
When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0
neonofarm [45]

Answer:

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Explanation:

mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625

mole of Na₂SO₄ = 2 x 0.025 = 0.05

                                      Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

( Mole/Stoichiometry )    \frac{0.0625}{1}           \frac{0.05}{1}

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From  (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.

Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄

                                            = 0.05 x 303.26 g

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