1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
jarptica [38.1K]
3 years ago
7

Convert 0.76 kg to grams. plz help, thank you

Chemistry
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

the second answer

Explanation:

just multiple 10²=10×10=100×7.6=760g

You might be interested in
For a certain reaction, Kc = 2.76×103 and kf= 6.54×10−4 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given th
riadik2000 [5.3K]

Answer:

2.37x10⁻⁷ M⁻².s⁻¹

Explanation:

For a generic reversible reaction:

A + B ⇄ C + D

Kf is the constant of the formation of the products (C and D), Kr is the constant of the formation of the reactants (A and B), and Kc is the general equilibrium constant, which is:

Kc = Kf/Kr

2.76x10³ = 6.54x10⁻⁴/Kr

Kr = 6.54x10⁻⁴/2.76x10³

Kr =  2.37x10⁻⁷ M⁻².s⁻¹

5 0
3 years ago
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

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:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

8 0
3 years ago
How does an increase in temperature affect a chemical reaction??
blsea [12.9K]
An increase in the temperature will speed up the reaction by increasing the frequency and efficiency of the collisions of molecules.
8 0
3 years ago
Which sample of matter is classified as a solution? 1. H2O(s) 2. H2O(l) 3. CO2(g) 4. CO2(aq)
lys-0071 [83]

Answer : Option 4) CO_{2}_{(aq)}

Explanation : CO_{2}_{(aq)} is the only sample of matter which can be classified as a solution. As the solution can be defined as a liquid mixture which contains a minor component (the solute) that is uniformly distributed within the major component (the solvent).

In this case, the solute is CO_{2} which is dissolved in water which acts as an solvent. Also, it has a subscript which is aq. which means aqueous, is often given to the solution in which the solvent is water.

Therefore, CO_{2}_{(aq)} is the correct answer.

4 0
3 years ago
Read 2 more answers
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o
sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4}  } = 0.0106 M

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

4 0
3 years ago
Other questions:
  • As the pressure on a sample of gas increases, the volume of the sample _____.
    13·2 answers
  • Polar substances will usually dissolve in _____.
    12·2 answers
  • Is celery organic chemically? If so, what is it (carbohydrate, protein, etc)?
    10·1 answer
  • A certain radioisotope has a half-life of 7.32 days. what percentage of an initial sample of this isotope remains after 36 days?
    12·1 answer
  • Select all (there's more than one) of the below chemicals that would be insoluble in water.
    9·1 answer
  • What is the average kinetic energy of 1 mole of gas at 0 degrees kelvin
    12·1 answer
  • Formula weight of H2CO3
    9·2 answers
  • How are mixtures and solutions different from each other?
    5·1 answer
  • What would you expect to observe when Br2 reacts with 2-butyne?
    14·1 answer
  • How many grams of sodium are in 10g. of aodium sulfate
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!