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lora16 [44]
3 years ago
7

What is another term for a Static Load?

Chemistry
2 answers:
Mamont248 [21]3 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]3 years ago
5 0

Answer: static charge

Explanation:

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A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph
Mashutka [201]
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
5 0
3 years ago
Iron (III) oxide pentahydrate formula
Nina [5.8K]

Answer:

Molecular Formula Fe2O12S3·5H2O

IUPAC Name iron(3+);tri sulfate;pentahydrate

Explanation:

4 0
3 years ago
A 1.250-g sample of benzoic acid, C7H6O2, was placed in a combustion bomb. The bomb was filled with an excess of oxygen at high
Degger [83]

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>

1.250g * (1mol / 122.12g) = 0.0102 moles

<em />

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

<em>3224 kJ/mol </em>

4 0
3 years ago
Please help.
Anna35 [415]

Answer:

9429.9

Explanation:

periodt

4 0
3 years ago
Given that you added about 5ug of purified acid phosphatase to tube A, calculate the amount of acid phosphatase that was present
Reil [10]

Hey there!:

Amount of purified acid phosphatase added to tube A = 5 ug ( micrograms )

Amount of acid phosphatase present in 400 ug of wheat germ extract  in tube B :

1 / 100 = 5x / 400 =

100 x = ( 0.5 ) ( 400 ) =

x = (0.5 ) ( 400 ) = 100

x = 200 / 100

x =  2 ug


The amount of acid phosphatase present in 400 ug of the  wheat germ in tube B is 2 ug


Hope that helps!

6 0
4 years ago
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