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2 years ago

14

what tool do you have to use to precise amount of water? A meter stick B stopwatch C balance D graduated cylinder E beaker F the

rmometer

1 answer:

Serhud [2]2 years ago

8 0

Answer:

graduated cylinder

Explanation:

It shows exact volume very precisely.

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I need help with these questions fast plz no links!!!!

nydimaria [60]

Answer:

c

B

B

I Hope u can understand and it's correct answer

8 0

2 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

SOMEBODY PLEASE HELP.. I DONT UNDERSTAND

elena-s [515]

Na,C12 is the answer

4 0

2 years ago

How many moles are present in 1.75 g of zinc sulfate, Znso, ?

Cerrena [4.2K]

Explanation:

We assume you are converting between grams ZnSO4 and mole. You can view more details on each measurement unit: molecular weight of ZnSO4 or mol This compound is also known as Zinc Sulfate. The SI base unit for amount of substance is the mole. 1 grams ZnSO4 is equal to 0.0061941519772353 mole.

7 0

3 years ago

1) A toy car with a mass of 2.35 kg is pushed with a force of 21.6 N. What is the acceleration of the car?

Nataly_w [17]

Answer:

The acceleration of the car is 9,19 m/s2

Explanation:

We use the formula: F=m x a---> a=F/m

a=21,6N/ 2,35kg 1N is 1kgxm/s2

a=21,6 kg x m/s2 x 2,35 kg

a=9,191489362 m/s2

6 0

3 years ago