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2 years ago

HELP ASAP

Chemistry

1 answer:

aksik [14]2 years ago

7 0

Answer:

C: Object 1 is moving, and Object 2 is not moving

Explanation:

Object one has unbalanced forces, so it can cause it to start moving ,since one force is moving stronger than its opposing one, but since Object 2 has balanced forces both forces hold it in place.

You might be interested in

0.25 g Al equals how many moles?

Mariulka [41]

0.25 g al should equal around 0.009265594867125 moles

I hope this help

5 0

3 years ago

Read 2 more answers

For chemical reactions involving ideal gases, the equilibrium constant K can be expressed either in terms of the concentrations

miskamm [114]

Answer:

$K_p= 3966.01$

Explanation:

The relation between Kp and Kc is given below:

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

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2CH_4_{(g)}\rightleftharpoons C_2H_2_{(g)}+3H_2_{(g)}$

Given: Kc = 0.140

Temperature = 1778 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (1778 + 273.15) K = 2051.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (3+1)-(2) = 2

Thus, Kp is:

$K_p= 0.140\times (0.082057\times 2051.15)^{2}$

$K_p= 3966.01$

6 0

3 years ago

A student carried out a titration using HC2H3O2(aq)HC2H3O2(aq) and NaOH(aq)NaOH(aq). The net ionic equation for the neutralizati

AnnZ [28]

Answer:

The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

Explanation:

Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:

CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O

The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio

Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M

Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles

Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles

Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5

There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.

Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

4 0

3 years ago

Plz help thx brainliest btw :)

Marina86 [1]

Answer:

1. earth metal 2. halogen 3.The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. ... Elements in the periodic table are organized according to their properties.

Inventor: Dmitri Mendeleev please name most brainliest

3 0

3 years ago

Read 2 more answers

I really can’t do this don’t understand it

masya89 [10]

Answer:

7.96g, 33.79%

Explanation:

I'll try my best to explain the entire process behind this question ;)

From the question, you can write the reaction

$2H_2O(l)->2H_2(g)+O_2(g)$

Now, there are a few reasons it is like this. Oxygen is a diatomic element, meaning it doesn't and can't exist as just O. It exists as O₂. To balance, this, double the amount of water and hydrogen so there is an equal amount of each element on both sides of the reaction (4 H's, 2 O's on the reactant side, and 4 H's, 2 O's on the product side).

From this we can get a mole-to-mole ratio.

Onto the stoichiometry. Our goal in this is to convert from grams of water to grams of hydrogen, and we do so with a mole to mole ratio.

$71.0gH_2O*\frac{1molH_2O}{18.02g} *\frac{2molH_2}{2molH_2O}* \frac{2.02g}{1molH_2}\\\\ =7.96gH_2$

Basically, what I did was divide by water's molar mass to get moles of water, multiplied by the mole-to-mole ratio (2:2) to get moles of H2, and then multiplied by H2's molar mass to get what should be the amount of H2 produced by the reaction.

For percent yield, you can calculate it is such:

$\frac{Actual}{Theoretical}*100$

Plug the numbers in:

$\frac{2.69g}{7.96g}*100\\\\ =33.79%$

So, the percent yield is 33.79%

3 0

3 years ago