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3 years ago

What chemical is discovered to react to light in the 1700s?

Chemistry

2 answers:

WARRIOR [948]3 years ago

5 0

When Scheele discovered oxygen he called it "fire air" as it supported combustion. The answer is oxygen.

leonid [27]3 years ago

4 0

Oxygen or maybe hydrogen

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What are some of the main ways to add energy to a reaction? plzzz help

Tresset [83]

Answer:

Heat.

Light.

Sound.

Electricity

Explanation:

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The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max

GuDViN [60]

Answer:

The maximum amount of copper allowed in 100 g of water is 0.00013 g

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg) and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

0.13 mg × 1 g / 1,000 mg = 0.00013 g

Answer: 0.00013 g of copper.

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3 years ago

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How many atoms are in 25.00 g of B.

iren2701 [21]

Answer:

$\boxed {\boxed {\sf 1.393 *10^{24} \ atoms \ B}}$

Explanation:

1. Convert Grams to Moles

Use the molar mass (found on the Periodic Table) to convert from grams to moles.

Boron (B): 10.81 g/mol

Use this value as a ratio.

$\frac {10.81 \ g \ B }{1 \ mol \ B}$

Multiply by the given number of grams.

$25.00 \ g \ B *\frac {10.81 \ g \ B }{1 \ mol \ B}$

Flip the ratio so the grams of boron cancel out.

$25.00 \ g \ B *\frac {1 \ mol \ B }{10.81 \ g \ B}$

$25.00 *\frac {1 \ mol \ B }{10.81 }$

$\frac {25.00 \ mol \ B }{10.81 }=2.312673451 \ mol \ B$

2. Convert Moles to Atoms

We use Avogadro's Number, 6.02*10²³: the number of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, the particles are atoms of boron.

$\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

Multiply by the number of moles we calculated.

$2.312673451 \ mol \ B *\frac {6.02*10^{23} \ atoms \ B} {1 \ mol \ B}$

The moles of boron cancel.

$2.312673451 *\frac {6.02*10^{23} \ atoms \ B} {1 }$

$2.312673451 *6.02*10^{23} \ atoms \ B} =1.39269195*10^{24} \ atoms \ B$

The original value of grams has 4 significant figures, so our answer should have the same. For the number we calculated, that is the thousandth place.

$1.392\underline69195*10^{24} \ atoms \ B$

The 6 tells us to round the 2 to a 3.

$1.393 *10^{24} \ atoms \ B$

25.00 grams of boron is equal to 1.393*10²⁴ atoms.

6 0

3 years ago