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SIZIF [17.4K]
3 years ago
10

Weight measures (2 points) how dense an object is. how large an object is. the amount of mass in an object. the effect of gravit

y on an object.
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
4 0

A weight of an object is a measure of the force exerted on the object by gravity. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.

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Methane burns in oxygen to produce carbon dioxide and water. Which of the following represents
Brut [27]

Answer:

A

Explanation:

CH4+O2-CO2+ H20

that mean methane has burn in oxygen to produce CO2

8 0
3 years ago
Study the solutions in the glasses. Put the solutions in order from concentrated to dilute.
Svetach [21]
A. 1,2,3. The solutions are getting lighter meaning the concentration is decreasing. Its most likely that water was added to dilute the solutions.
8 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
Consider a cell membrane at 298K which has glucose at the concentration of 200mM inside the cell and 20 mM outside the cell, wit
OleMash [197]

Answer:

D. +5.7 kJ/mol

Explanation:

Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:

ΔG = RT ln C in / C out

knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM

ΔG = 5,7 kJ/mol

Right answer is:

D. +5.7 kJ/mol

I hope it helps!

4 0
3 years ago
Which equation is NOT balanced?
lakkis [162]

Answer:

C

Explanation

On the reactants side there is 4 Hydrogen atoms in total and two oxygen atoms on the left however on the right there is two hydrogen atoms and one oxygen atom. Leaving this equation unbalanced

3 0
2 years ago
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