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3 years ago

10

Which one of the following temperatures is equal to 5°C?

1 answer:

natali 33 [55]3 years ago

7 0

Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature $5^oC$ .

Now the conversion factor used for the temperature is,

$K=^oC+273$

where, K is kelvin and $^oC$ is Celsius.

Now put the value of temperature, we get

$K=5^oC+273=278K$

Therefore, the temperature 278 K is equal to the $5^oC$

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Which star listed is the brightest intrinsically? Which is the intrinsically faintest?

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Answer Measurement. The Sun is the brightest star as viewed from Earth, at −26.74 mag. The second brightest is Sirius at −1.46 mag.

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3 years ago

Select all the correct answers.

WARRIOR [948]

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3 years ago

A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w

barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

$Moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}$

$Moles\ of\ AgBr= 0.1225\ mol$

The reaction taking place is:

$MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4$

From the reaction,

4 moles of AgBr is produced when 1 mole of $MBr_4$ undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of $MBr_4$ undergoes reaction

0.1225 mole of AgBr is produced when $\frac {1}{4}\times 0.1225$ mole of $MBr_4$ undergoes reaction

Moles of $MBr_4$ got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of $MBr_4$ = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

$0.030625 = \frac{12.5843}{319.616 + x}$

Solve for x,

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<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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3 years ago

All matter is made of _______________, which are made of ________________, __________________, and _____________________.

bonufazy [111]

All matter is made of atoms, which is made of protons, neutrons and electrons. Shouldn't know what else it is

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3 years ago