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kvv77 [185]
3 years ago
13

How many grams of Chromium can be formed when 150 grams of

Chemistry
1 answer:
valentinak56 [21]3 years ago
6 0

Answer: 104 g

Explanation: reaction Cr2O3 + 3 H2 ⇒ 2 Cr + 3 H2O

M(Cr2O3) = 150 g/mol, so n = m/M = 1.0 mol

Number of moles of H2 should be 3.0 moles and

It is much greater (150 g / 2.016 g/mol)

1 mol Cr2O3 produces 2 mol Cr.

Mass m= 2.0 mol· 52g/mol= 104 g

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Aloiza [94]

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Frequency is 2Hz

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1 year ago
True or False: Dishwashing liquid cleans dirty plates because the nonpolar tail of the detergent dissolves the oily dirt particl
Ghella [55]

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true

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3 0
2 years ago
When molten candle wax solidifies is that a chemical reaction or not?l
andrey2020 [161]

Yes, when molten candle wax solidifies it is a chemical reaction

<u>Explanation:</u>

Basically Wax is crystalline so once the candle light melts it freezes taking back the solid state to the room temperature.

When the room temperature is below the freezing point, the liquid candle wax, turns into solid state again, therefore this process is called solidification. The process of freezing or solidification is a process when an object turns liquid and freezes back to solid state.

Indeed, Yes, when molten candle wax solidifies it is a chemical reaction

3 0
3 years ago
The prefix 'kilo-' means a thousand. So a kilo-gram is a thousand grams. 1 kilogram=1,000 grams How many kilograms is 1,008 gram
lesantik [10]

<u>Answer:</u>  1.0 kilograms.

<u>Explanation:</u>

One kilogram is equal to a thousand grams.

Supposing x to be the number of kilograms equal to one thousand and eight grams, we can write it as:

1 kg = 1000 grams

x kg = 1008 grams

To solve for x, we can simply divide 1008 grams by 1000 thousand grams to get the answer.

x = 1008 / 1000

x = 1.008

Rounding this value to the nearest tenth, it will become 1.0 kilograms.

4 0
3 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
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