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lakkis [162]
3 years ago
5

The answer to #21 , anybody ??

Chemistry
2 answers:
sergij07 [2.7K]3 years ago
8 0
The answer to your question is c
Temka [501]3 years ago
4 0
Answer:

C)

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True or False: A nail with a wire wrapped around it is<br> an electromagnet.
Annette [7]

Answer:Search ResultsBy simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

Explanation:

7 0
3 years ago
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Is bleach liquid starch? <br> Yes or No
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O it's not

Explanation:

Have a great day!

8 0
2 years ago
Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?
Setler79 [48]

Answer:

\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

  • molarity= 1.2 mol NaNO₃/L
  • liters of solution=0.25 L
  • moles of solute =x

1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L

0.25 \ L *1.2 \ mol \ NaNO_3/L=x

The units of liters cancel, so we are left with the units moles of sodium nitrate.

0.25  *1.2 \ mol \ NaNO_3=x

0.3 \ mol \ NaNO_3=x

There are 0.3 moles of sodium nitrate.

3 0
2 years ago
How many valence electrons are in each atom? a. potassium C. magnesium b. carbon d. oxygen ​
skad [1K]
Just look it up you’ll get the answers faster
7 0
2 years ago
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
3 years ago
Read 2 more answers
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