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3 years ago

Students investigate forces and how they affect objects. They use a small car and wooden blocks to carry out their investigation

Chemistry

1 answer:

Umnica [9.8K]3 years ago

7 0

Answer:

increase the number of blocks

increase the number of students pulling the cart

increase the number of blocks and increase the number of students pulling the cart

decrease the number of blocks

decrease the number of students pulling the cart

decrease the number of blocks and decrease the number of students pulling the cart

(ANSWER CHOICES)

Explanation:

You might be interested in

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 23 g of acetic acid, CH3COOH, and dissolving

Elza [17]

This is an incomplete question, here is a complete question.

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH₃COO, to 23 g of acetic acid, CH₃COOH, and dissolving in water to make 400. mL of solution. Hint given in feedback. The Ka for CH₃COOH is 1.8 x 10⁻⁵ M. As usual, report pH to 2 decimal places.

Answer : The pH of the solution is, 4.97

Explanation :

First we have to calculate the moles of CH₃COOH and NaCH₃COO

$\text{Moles of }CH_3COOH=\frac{\text{Given mass }CH_3COOH}{\text{Molar mass }CH_3COOH}=\frac{23g}{60g/mol}=0.383mol$

and,

$\text{Moles of }CH_3COONa=\frac{\text{Given mass }CH_3COONa}{\text{Molar mass }CH_3COOH}=\frac{59g}{82g/mol}=0.719mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.7$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}$

Now put all the given values in this expression, we get:

$pH=4.7+\log (\frac{0.719}{0.383})$

(As the volume is same. So, we can write concentration in terms of moles.)

$pH=4.97$

Therefore, the pH of the solution is, 4.97

6 0

4 years ago

What is the relationship between grams and cm³?

soldier1979 [14.2K]

Answer:

1 gram

Explanation:

1 gram= weight of 1 cubic centimeter of pure water at temperature 4°C = 1000 milligrams = 0.001 kilogram

3 0

3 years ago

Consider the following chemical equilibrium: 4NH3+3O2=2N2+6H2O Now write an equation below that shows how to calculate from for

erma4kov [3.2K]

This is an incomplete question, here is a complete question.

Consider the following chemical equilibrium:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Now write an equation below that shows how to calculate Kp from Kc for this reaction at an absolute temperature T. You can assume T is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.

Answer :

The expression of $K_c$ will be,

$K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}$

The expression of $K_p$ will be,

$K_p=K_c(RT)$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

The expression of $K_c$ will be,

$K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}$

The relation between $K_p$ and $K_c$ is:

$K_p=K_c(RT)^{\Delta n}$