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Sergeeva-Olga [200]
3 years ago
11

Please help i will mark brainlist

Chemistry
2 answers:
Sphinxa [80]3 years ago
4 0

Answer:

V= 21 cubic cm (( I am not sure about density ))

Explanation:

V= L×W×H

Elan Coil [88]3 years ago
4 0

Answer:

volume = 21 cubic cm or 21 cm^3

density = 3 g/cm^3

Explanation:

volume = length × breadth × height

  • vol = 7 × 3 × 1 = 21 cubic cm

hence, volume = 21 cubic cm.

density = mass / volume

  • density = 63 / 21 = 3 g / cm^3

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>. </em><em>.</em><em>.</em><em>.</em>

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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If a lab requires each lab group to have 25ml of a solution and it takes 15 grams of CuNO3 to make 1 liter of solution how many
tino4ka555 [31]

We need to do some general algebra here.

We will find that you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

<em>We know that:</em>

  • Each lab group needs 25 ml of solution.
  • it takes 15 g of CuNO₃ to make one L of that solution.
  • There are 22 labs.

Because each lab needs 25 ml of solution, 22 labs will need that amount 22 times, so the <u>total amount of solution needed</u> is:

22*25ml = 550 ml

Now we know that we need 15 grams to make one liter of solution, and:

1 L = 1000ml

Then you need 15g to make 1000ml

and x (we want to find this amount) to make 550ml

Then we can write two equations (not actual equations, as these are different units) like:

x = 550ml

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Now we can take the quotient between these two equations:

x/15 g = (550ml/1000ml)

And now we can solve this for x:

x = (550ml/1000ml)*15g = 8.25g

So you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

If you want to learn more, you can read:

brainly.com/question/8743486

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