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Anton [14]
3 years ago
7

An apple has a mass of 297 grams. How many kilograms is this? (Note: You can use the information in the table below to make a co

nversion factor.)
Chemistry
2 answers:
Fantom [35]3 years ago
7 0

Answer: 0.297 grams.

Explanation: the conversion factor between kilograms and grams is:

1 kg=1000 g

applying this factor to the given mass of an apple, we have:

297 grams *\frac{1 kg}{1000 grams}

as we can see, we just have to divide by 100 so:

297 grams= 0.297 kg.

yan [13]3 years ago
4 0
.297 kilograms
297g x 1 kg/100 g = .297
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What is the phase change in which a solid turns into a liquid
STatiana [176]

Answer:

melting

Explanation:

5 0
3 years ago
if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
3 years ago
Find the volume of a gas if 3.6 mols of the gas is at STP
Digiron [165]

Answer:

81 L gas

General Formulas and Concepts:

<u>Ideal Gas Law</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.6 mols gas at STP

[Solve] volume (L) of gas

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.6 \ mol \ gas(\frac{22.4 \ L \ gas}{1 \ mol \ gas \ at \ STP})
  2. [DA] Multiply [Cancel out units]:                                                                      \displaystyle 80.64 \ L \ gas

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

80.64 L gas ≈ 81 L gas

6 0
3 years ago
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3 . How could you determine whether the solution is
navik [9.2K]

Explanation:

A solution is said to saturated when it cannot dissolve any extra solute in it. The extra solute put remains undissolved.

A solution is said to unsaturated, when the concentration of solute is less as compared to solubility of the solution it is said to be unsaturated.

A solution is said to be super saturated when it contains more of the solute than the solvent  can dissolve under normal conditions is called super saturated.

5 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
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