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ki77a [65]
3 years ago
11

A little bigger than the diameter of the cylinder, _________ stops pressurized fuel air mixture from leaking between the cylinde

r wall and the pistons
Chemistry
1 answer:
Vladimir [108]3 years ago
7 0

The correct term here is a "piston ring."

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How many grams of hcl would be produced if 54 grams of water were used
katen-ka-za [31]

Answer:

i dont know

Explanation:

its simple i dont know

3 0
3 years ago
Read 2 more answers
Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent
elixir [45]

Answer : The initial rate for a reaction will be 3.8\times 10^{-4}Ms^{-1}

Explanation :

Rate law : It 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.

The chemical equation will be:

A+B+C\rightarrow P

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:

6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^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(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^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:

6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1

k=7.6\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

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

\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1

\text{Rate}=3.8\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.8\times 10^{-3}Ms^{-1}

6 0
3 years ago
Consider the following balanced equation:
Serga [27]
<span>Zn⁰ + 2H⁺ ------> Zn²⁺ + H2⁰

H⁺ ion has oxidation number +1.
Zn²⁺ ion has oxidation number +2.
Atom of Zn has electric charge 0, and each hydrogen atom in the molecule H2 have oxidation number 0. So, </span> Zn and each hydrogen atom in H2 have oxidation numbers equals "0".<span>

</span><span>Answer is
D. Zn and each hydrogen atom in H2</span><span>

</span>
4 0
3 years ago
Name all 7 elements classified as metalloids. <br> PLEASE HELP ASAP
Strike441 [17]

Answer:

Explanation

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po

6 0
3 years ago
Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP
DIA [1.3K]

The second volume :    V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :  

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0
3 years ago
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