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mart [117]
3 years ago
11

Match each lab incident to the item of PPE that will protect you from it. Each item of PPE will only be used once.

Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

fefeff

Explanation:

wdwdwfefe

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Is gunpowder a homogenous mixture or a heterogenous mixture? Explain why.
inysia [295]

Gunpowder is a heterogenous mixture,

Reason : the composiotion of the mixture is not uniform throughout, the mixture can be separated by a suitable method , and finally , it composes of charcoal, sulfur and potassium nitrate which is not homogeneous mixture.

4 0
1 year ago
Who developed the orbital model of the atom?
Nataliya [291]

Answer:

Ernest Rutherford

Explanation:

that's I think

7 0
3 years ago
What would the formula be for this model above?
Nadusha1986 [10]
Probably CH(subscript)4... :) It's Methane
4 0
3 years ago
How many grams of oxygen gas (02) are needed to completely react with 9.30 moles
Luba_88 [7]

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^         ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al          3 moles O₂              32.0 g
-------------------  x  ---------------------  x  --------------------  =  223 g O₂
                              4 moles Al               1 mole

6 0
2 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
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