HELP ME PLEASE I BEG YOU

Which describes the slope of the graph from 10 s to 30 s?

inna [77]

Answer:

3

Explanation:

Use the rise over run method. plot the necessary points first in the x and y axis

6 0

3 years ago

Which of the following are the starting substances in photosynthesis?

Misha Larkins [42]

A: water and carbon dioxide.

3 0

3 years ago

Read 2 more answers

Nitrifying bacteria participate in the nitrogen cycle mainly by

Brilliant_brown [7]

Hello, here’s the answer to your question. Converting ammonia to nitrate, which is absorbed by plants

6 0

3 years ago

4. How many grams of ammonium carbonate are needed to decompose in order to produce

Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0

3 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 $C_{1}H_{2}=CH_2$

4 0

3 years ago