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2 years ago

What is another term for a Static Load?

Chemistry

2 answers:

Mamont248 [21]2 years ago

7 0

Answer:

Explanation:

static load, also called the holding load, is the force that will be applied to the linear actuator when it is not in motion.

Ronch [10]2 years ago

5 0

Answer: static charge

Explanation:

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When you exercise strenuously, your body produces excess heat. Describe what your body does to help prevent your temperature fro

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Answer:

Your body can cool itself by sweating. When sweat evaporates, it lowers your temperature

Explanation:

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2 years ago

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If you were preparing a defined, minimal medium for an aerobic, chemolithotrophic bacterium to grow on, which of the following c

Eddi Din [679]

Answer:

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Explanation:

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2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

18. Which metal is capable of forming more than one cation? Li Ba Al Sn

maw [93]

Answer:

Explanation:

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3 years ago

PLEASE HELPPPP???!!!!!!

sertanlavr [38]

Mass of molecule (g) = Mr of substance over avarogado constant

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3 years ago