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docker41 [41]
3 years ago
12

What force must act on a 150 kg mass to give it an acceleration of 30 m/s squared? F=ma

Chemistry
1 answer:
Helga [31]3 years ago
4 0

Answer:

its 120

Explanation:

i did it :)

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Please help! Best answer will receive brainliest. 15 points.
9966 [12]

Answer:

Melting of ice is a physical change and during this change latent heat of fusion, that is, heat required to convert 1 g of a solid substance into liquid state will keep the temperature constant unless and until all the solid changes into liquid. On the other hand, a change in which a new product is formed due to change in chemical composition of the reacting species is known as a chemical change. During a chemical change, there will occur change in temperature. So, when a substance is set on fire then there will occur change in temperature of the substance. Hence, it means it is a chemical change.

Explanation:

Let me know if you need more clarification!!!

3 0
4 years ago
Determine how many liters 8.65 g of carbon dioxide gas would occupy at the following conditions.
VashaNatasha [74]
1) 6.17 liters

2)2.66 liters
3 0
3 years ago
How many atoms are present in 0.035 moles?
RUDIKE [14]

Answer:

2.11x10^{22}atoms

Explanation:

Hello!

In this case, considering the Avogadro's number, which relates the number of particles and one mole, we can infer that 1 mole of any element contains 6.022x10²³ atoms as shown below:

\frac{6.022x10^{23}atoms}{1mol}

Thus, we compute the number of atoms in 0.035 moles as shown below:

atoms=0.035mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms=2.11x10^{22}atoms

Best regards!

7 0
3 years ago
What volume (in mL ) of 0.200   M NaOH do we need to titrate 40.00 mL of 0.140   M HBr to the equivalence point?
Ugo [173]
NaOH + HBr =⇒ NaBr + H2O

35.0 ml HBr x 1 liter/1000 mL x 0.140 moles HBr/ liter = 0.0049 moles HBr

0.0049 moles HBr x 1mole NaOH/1mole HBr = 0.0049 moles HBr

0.0049 moles HBr x 1 liter NaOH/0.200 moles NaOH x 1000 mL/1liter= 24.5 mL NaOH
7 0
3 years ago
Read 2 more answers
Calculate the PH of 0.25mol H2SO4
Hoochie [10]

<span>This problem uses the relationship between Ka and the concentrations of the ions. Calculations are as follows:</span>

<span>
</span><span>1.9 x 10-5</span>= x^2 / (0.25 - x)

<span>x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3

</span><span>pH = -log (2.2 x 10^-3)</span> = 2.66

7 0
3 years ago
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