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2 years ago

1. Each substance written to the right of the arrow in a chemical equation is a

Chemistry

1 answer:

Scorpion4ik [409]2 years ago

5 0

Answer: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

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At a certain temperature the speeds of six gaseous molecules in a container are 2.0 m/s, 2.2 m/s, 2.6 m/s, 2.7 m/s, 3.3 m/s, and

iren [92.7K]

Answer:

root-mean-sqaure = 2.77 m/s

average = 2.72 m/s

The root-mean-square is always the largest because it takes account of the variance of the spread of the data. The increase is related to the fact that the data varies to sample.

Explanation:

The rootmean-square (R) is the square root of the squares of the valeus divided by the number of the datas.

$R = \sqrt{\frac{x1^2 + x2^2 +...+xn^2}{n} }$

$R = \sqrt{\frac{(2.0)^2 + (2.2)^2 + (2.6)^2 + (2.7)^2 ^(3.3)^2 + (3.5)^2}{6} }$

R = √(46.03)/6

R = 2.77 m/s

The average speed is the sum of the speeds divided by the number of datas:

$A = \frac{2.0 + 2.2 + 2.6 + 2.7 + 3.3 + 3.5}{6}$

A = 16.3/6

A = 2.72 m/s

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3 years ago

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You have a volumetric flask containing 0.54 M HBr. The SOLUTE is HBr and the SOLVENT is water.

soldi70 [24.7K]

Answer:

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Explanation:

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2 years ago

A reaction between substances Y and Z is

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Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

$K=A\times e^{\frac{-Ea}{RT}}$

Taking logarithm on both the sides, we get:

$\ln k=-\frac{Ea}{RT}+\ln A$ ............(1)

where,

k = rate constant = $2.95\times 10^{-3}L/mol.s$

Ea = activation energy = ?

T = temperature = 435 K

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor = $3.00\times 10^{+10}L/mol.s$

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

$\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)$

$Ea=108318.365J/mol=108.318kJ/mol$

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

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Theresa invested $5,000 in an account she expects will earn 7% annually. Approximately how many years will it take for the accou

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I: Current
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