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3 years ago

A base is a substance that's the chemical opposite of an acid. True or false

Physics

1 answer:

anygoal [31]3 years ago

6 0

True

Explanation:

A base is a substance that is often used as the chemical opposite of an acid.

Both behaves in opposite way to one another.

They can be said to complementary or conjugate chemicals.

According to Bronsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor.
The lewis theory states that an acid is an electron pair acceptor while a base is an electron pair donor.
Acids turn blue litmus paper red and bases turns red litmus paper blue.

learn more :

Acid brainly.com/question/11062486

#learnwithbrainly

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The spans of time that make up the Earth's geologic timescale are generally defined by the different types of

Trava [24]

division of Earth's history into time units based largely on the types of life-forms that lived only during certain periods.

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3 years ago

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453

Drupady [299]

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

$0.0001\times \frac{100}{0.4539}=0.022%$

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

$\frac{1}{0.022}\times {100}=4545.5\ kg$

Mass in pounds would be

$\frac{4545.5}{0.4539}=10014.32\ lb$

Mass in pound-mass is 10014.32 lb

8 0

3 years ago

Help please asap I'm super lost

vaieri [72.5K]

A compound is two or more elements combinded together.
A element is a pure substance.
it looks like all of them are compounds except the carbon dixode, My teacher would accept that as a anwser idk bout urs thoughs.

6 0

2 years ago

Read 2 more answers

A candle is placed 30 cm in front of a convex mirror with a focal length of 20 cm, as shown in the diagram. What is the distance

lora16 [44]

I think the answer is 60 cm.

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3 years ago

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago