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olasank [31]
3 years ago
10

Express the numbers 0.00087 and 54,200,000 in scientific notation

Chemistry
2 answers:
Slav-nsk [51]3 years ago
6 0

Answer: a) 8.7\times 10^{-4}

b) 5.42\times 10^{6}

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as 5.0\times 10^3

a) 0.00087

This number is written in scientific notation as : 8.7\times 10^{-4}

b) 54,200,00

This number is written in scientific notation as : 5.42\times 10^{6}

yuradex [85]3 years ago
5 0
8.7 x 10^-4
5.42 x 10^7
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If an element has a mass number of 222 and an atomic number of 86, how
svlad2 [7]

Answer:

136

Explanation:

The Mass Number is the combination of the amount of Protons and Neutrons in an element, so if the total mass is 222, and the amount of protons is 86, then you can do 86 + x = 222 to find that x is equal to 136

5 0
2 years ago
Please Help! It says fill in the blanks to complete the Punnett square below to show the results of a cross between a homozygous
makkiz [27]

Answer:

Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.

Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%

Explanation:

Hello,

The Punnett square for this cross turns into:

\left[\begin{array}{ccc}&G&g\\G&GG&Gg\\G&GG&Gg\end{array}\right]

It means that the genotypes and phenotypes are:

Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.

Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%

Best regards.

4 0
2 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
What is the total energy change for the following reaction:CO+H2O-CO2+H2
Alekssandra [29.7K]

Answer:

\large \boxed{\text{-41.2 kJ/mol}}

Explanation:

Balanced equation:    CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

4 0
2 years ago
A commercial airplane typically flies at a height of 37,000 feet. If an average cell phone was dropped from an airplane, about h
saul85 [17]

Answer:

its b i just did it .

6 0
3 years ago
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