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3 years ago

I smell onions.

Chemistry

2 answers:

goldenfox [79]3 years ago

5 0

Answer:

Observation

-I smell onions

-My dog weighs 45 pounds

-It is 85 degrees outside today

Predictions

-My dog will bark at the vacuum cleaner

although he has never seen it before

-My mother is going to cook spaghetti for

supper next Tuesday

-It will be a long, hot summer

Explanation:

ioda3 years ago

3 0

Observation: I smell onions, it’s 85°F outside today, my dog weighs 45 pounds,

Prediction: it will be a long hot summer
My dog wil bark... , my mother...

(I am not 100% sure)

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1 gallon is how many mm

leva [86]

Answer:

0.133681 (cubic foot)

Explanation:

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2 years ago

Which one of the following is different from the others A .HCl B. HF C. HBr D. HI

lora16 [44]

Answer:

D.HI

Explanation:

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3 years ago

You have 16.7 grams of hydrogen and 15.4 grams of oxygen in a synthesis rxn. Which is the limiting reagent?

sleet_krkn [62]

Answer:

oxygen is limiting reactant

Explanation:

Given data:

Mass of hydrogen = 16.7 g

Mass of oxygen = 15.4 g

Limiting reactant = ?

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 16.7 g/ 2 g/mol

Number of moles = 8.35 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 15.4 g/ 32 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of both reactant with product,

H₂ : H₂O

2 : 2

8.35 : 8.35

O₂ : H₂O

1 : 2

0.48 : 2×0.48 = 0.96 mol

The number of moles of water produced by oxygen are less so it will limiting reactant.

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3 years ago

An unknown element, X, reacts with sodium to form the compound Na2X. In other compounds, this element also can accommodate up to

Andreas93 [3]

Answer:

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3 years ago

What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of

Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 = -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m= molality

molality = $\frac{moles of solute}{mass of solvent in kilograms}$

putting the value in the equation:

molality= $\frac{1.31}{0.53}$

= 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

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3 years ago