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lyudmila [28]
2 years ago
13

20 Infer Check the box to show which would tell you for sure if you had a sample of real gold.

Chemistry
1 answer:
Llana [10]2 years ago
4 0

Answer:

C but read the limitation

Explanation:

Gold sample

Color wouldn't tell you much. In fact it may fool you. Fool's Gold is the same color as gold.

Gold is very malleable. That's what hitting gold with a hammer will do.  It tends to shape the gold without destroying it into lots of pieces. A lot of yellow things have that property.

If only it was that simply. Yes if you are sure there is gold in that location. You have to pick this answer, because the other 2 don't work. So this is possible, but not the best way. The best way is to subject the sample to chemical treatment. If you are left with a yellow sample, then you could make the case for it being gold.

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A car traveling in a northeasterly direction at a rate of 50 km/h represents which of the following?
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B. velocity

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Velocity is defined as speed in a given direction.

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3 years ago
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Answer:You can view more details on each measurement unit: molecular weight of BaCl2 or grams This compound is also known as Barium Chloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles BaCl2, or 208.233 grams.

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One person is pulling with a force of 250 newtons and the other person is pulling with a force of 200 newtons. The total force o
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3 years ago
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
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Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

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Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

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Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

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