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3 years ago

Pls help this is due now

Chemistry

2 answers:

Luba_88 [7]3 years ago

4 0

Answer:

B, Sleeping

Hope this helps :)

Mandarinka [93]3 years ago

3 0

Answer:

Sleeping

Explanation:

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‼️✨ HELP IMMEDIATELY 10 POINTS ‼️✨

cluponka [151]

Answer:

1. A

2. A

3. D

4. A

Explanation:

I just did this and these were the answers :)

3 0

4 years ago

The valency of metal 'm' 1+. give the formula of its oxide

Vinvika [58]

Answer:

$M_{2}O$

Explanation:

Anything oxide is a compound with oxygen, and since oxygen is -2, it requires two of the metal's +1 to make it zero

In other words:

2(+1) + (-1) = 0

7 0

3 years ago

Cooling a gas in a rigid container causes

maxonik [38]

Answer:

Explanation:

Presumably the system is closed (nothing can escape or enter).

A: You're cooling the container so it shouldn't explode.

C: If the container is Closed, the volume cannot increase or decrease.

D: is also impossible for the closed container reason.

B: The pressure has to decrease because the molecules are moving slower which decreases their velocity. If that happens the Force goes down.

B: Answer

4 0

3 years ago

How many microliters are in 7.4 x 10^-61 centimeters?

schepotkina [342]

Answer:

Microliters = $7.4\times 10^{-58}$

Explanation:

SI unit of volume is liter.

$1 dm^{3}=1 liter$ ..............................(1)

1dm=10cm

$1dm^{3} = 1000 cm^{3}$ .................(2)

replacing value of $1 dm^{3}$ from equation (1) in equation (2)

$1 liter = 1000 cm^{3}$ ...................(3)

but

$1 liter = 10^{6} microliters$ ....................(4)

replacing value of 1 liter from equation (3) in equation (4)

$1000 cm^{3} = 10^{6} microliters$

on solving further ,we get

$1 cm^{3} = \frac{10^{6}}{10^{3}}microliter$

$1 cm^{3} = 1000 microliter$

$7.4\times 10^{-61}\times 1000$

Microliters = $7.4\times 10^{-58}$

8 0

3 years ago

Calculate the number of joules of heat energy needed to increase the temperature of 25.0 g of metal from 21.0 ºC to 80.0 ºC. The

Harman [31]

Answer:

Q = 768.47 J

Explanation:

Given that,

Mass of the metal, m = 25 g

Initial temperature, T₁ = 21.0 ºC

Final temperature, T₂ = 80.0 ºC

The specific heat of the metal is 0.521 J/gºC.

We know that the heat released due to the change in temperature is given by :

$Q=mc\Delta T\\\\=25\times 0.521\times (80-21)\\Q=768.47\ J$

Hence, 768.47 J of heat energy will be needed.

7 0

3 years ago