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uranmaximum [27]
2 years ago
9

I need help asap!!!!!!!!

Chemistry
1 answer:
maksim [4K]2 years ago
6 0

Answer:

A

Explanation:

add 22 with the answers like A, 33+22=55

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If aluminum has a density of 2.7 g/cm³, what is the volume of 2.7 grams of aluminum?​
Lorico [155]

Answer:

Solution Density of aluminium = 2.7 g/Cm 3 In kg/ m 3 = 27 × 1000 10 =2700 kg/ m 3

Explanation:

Not much of one

4 0
1 year ago
A pet owner dropped some new fish into a pond. Several species of fish lived in the pond, but within a few months, only members
Daniel [21]

Answer: the correct answer is : too many fish caused an imbalance in the ecosystem, so the food web was affected.

Explanation:

8 0
3 years ago
Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9
Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid C_2H_5OH=-135^oC=138 K(0^oC=273K)

Melting temperature of Solid C_2H_5OH=114^oC=159 K

Temperature of liquid C_2H_5OH=-50^oC=223K

Specific heats of solid  ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid C_2H_5OH at 159 K

\Delta T_1 = 159 K - 138 K = 21 K

Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J

Heat required to melt and raise the temperature of C_2H_5OH upto 223 K

\Delta T_2 = 223 K - 159 K  = 64 K

Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J

Total heat to convert solid ethanol to liquid ethanol at given temperature :

Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0
3 years ago
1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?
tatyana61 [14]
<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

  • 10.0 M HNO₃

Prepared solution;

  • Volume of solution as 0.350 L
  • Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

  • We are going to use the dilution formula;
  • The dilution formula is;

M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

    =(0.40 M × 0.350 L) ÷ 10.0 M

   = 0.014 L

But, 1 L = 1000 mL

Therefore,

Volume = 14 mL

Thus, the volume of 10.0 M HNO₃ is 14 mL

5 0
3 years ago
If I have 5,000g of sodium, how much sodium sulfide can be made?
arlik [135]

2.469 could  be maid dont forget to hit that thanks button

8 0
3 years ago
Read 2 more answers
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