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Blizzard [7]
3 years ago
5

Really need help with the rest of them, please it's due in 20 minutes

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

chlorine is for pool

Explanation:

hope is help

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Find the number of moles of sodium hydroxide in 25cm3 solution of concentration 0,1 mol/dm3​
kozerog [31]

Answer:

0.0025  moles

Explanation:

25cm^3 = 25/1000 dm^3

Conc = mol/dm^3

Mol = conc * dm^3

Mol = 0.1 * 25/1000

3 0
2 years ago
If the concentration of a KCl solution is 16.0% (m/v), then the mass of KCl in 26.0 mL of solution is ________.
melomori [17]

Answer:

The correct answer is 4.16 grams.

Explanation:

Based on the given information, the concentration of KCl solution is 16 % m/v, which means that 100 ml of the solution will contain 16 grams of KCl.

The molarity of the solution can be determined by using the formula,

M = weight/molecular mass × 1000/Volume

The molecular mass of KCl is 74.6 grams per mole.

M = 16/74.6 × 1000/100

M = 16/74.6

M = 2.14 M

Now the weight of KCl present in the solution of 26 ml will be,

2.14 = Wt./74.6 × 1000 /26

Wt. = 4.16 grams

3 0
2 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
3 years ago
Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.
OlgaM077 [116]

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

3 0
3 years ago
Read 2 more answers
What is the pressure exerted by a 0.500 mol sample of nitrogen in a 10.0 L container at 20°C?
lidiya [134]
The answer would be D.120 kPa
6 0
3 years ago
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