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3 years ago

4. The vet instructed Manuel to give his

Chemistry

1 answer:

djverab [1.8K]3 years ago

3 0

Answer:

=60 milligrams

Explanation:

12 x 5

=60 milligrams

Have a nice day!!!!!!! :-)

KA

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Does wasting water also waste energy? Explain your reasoning.

77julia77 [94]

Answer:

Yes.

Explanation:

Wasting household water does not ultimately remove that water from the global water cycle, but it does remove it from the portion of the water cycle that is readily accessible and usable by humans. Also, "wasting" water wastes the energy and resources that were used to process and deliver the water.

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3 years ago

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Hydrogen produced from a hydrolysis reaction was collected over water and the following data was compiled.

Shkiper50 [21]

Answer:

0.00358 mol

Explanation:

1) Data:

a) V = 93.90 ml

b) T = 28°C

c) P₁ = 744 mmHg

d) P₂ = 28.25 mmHg

d) n = ?

2) Conversion of units

a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter

b) T = 28°C = 28 + 273.15 K = 301.15 K

c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm

d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm

3) Chemical principles and formulae

a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.

b) Ideal gas equation: pV = nRT

4) Solution:

a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm

b) Moles of hygrogen gas:

pV = nRT ⇒ n = pV / (RT) =

n = (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =

n = 0.00358 mol (which is rounded to 3 significant figures) ← answer

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3 years ago

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The decay of a living things allows chemical elements to be lost.

Lelu [443]

Answer: is is true

Explanation:

8 0

2 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: 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 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

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4 years ago

Andrej [43]

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3 years ago

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