PLEASE HELP ME!!! It would be great

Bees help flowers reproduce through a process called __________. *

Ostrovityanka [42]

Answer:

Pollination

Explanation:

5 0

4 years ago

Write functions of the following parts of the digestive system.

Lapatulllka [165]

Answer:

1. absorb water,get rid of of waste products

2.chewing

3. absorbing nutrients

4. holding the food, breakdown of food

3 0

3 years ago

Two substances A and B react with each other in such a way that one-half of A remains after 25 min and one-fourth of A remains a

juin [17]

Answer:

a. first-order in both A and B.

Explanation:

In order to determine the kinetics of reaction let us consider the given data

a) Doubling the concentration of B while keeping the concentration of A fixed doubles the rate of the reaction. : it means the reaction is first order reaction with respect to reactant B.

b) The half life is the time required to complete the reaction by half way.

As with time one half of A remains and after equal time one fourth of A remains it means again the order of reaction is first with respect to A also.

There is no effect on the half life if we are decreasing the initial concentration of A also.

3 0

3 years ago

When aqueous solutions of K2CO3 and Cu(NO3)2 are mixed, what precipitate(s) will form?

Mnenie [13.5K]

When aqueous solutions of Potassium Carbonate and Copper Nitrate are mixed, the potassium will displace the copper, as it is more reactive, and Potassium Nitrate and Copper Carbonate will be formed.

The equation for this reaction is (ignoring state symbols):

K₂CO₃ + Cu(NO₃)₂ ⇒ 2KNO₃ + CuCO₃

7 0

3 years ago

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B

Alborosie

Answer: The empirical formula for the given compound is $C_{15}H_{24}O_1$

Explanation:

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

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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

We are given:

Mass of $CO_2=5.995g$

Mass of $H_2O=1.963g$

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 5.995 g of carbon dioxide, $\frac{12}{44}\times 5.995=1.635g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.963 g of water, $\frac{2}{18}\times 1.963=0.218g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles$

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 0.0092 moles.

For Carbon = $\frac{0.136}{0.0092}=14.78\approx 15$

For Hydrogen = $\frac{0.218}{0.0092}=23.69\approx 24$

For Oxygen = $\frac{0.0092}{0.0092}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is $C_{15}H_{24}O_1$

3 0

4 years ago