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Mazyrski [523]
4 years ago
15

A basketball weighs approximately 1.35 kg. What is this mass in grams (g)? step by step

Chemistry
1 answer:
umka21 [38]4 years ago
5 0

Answer:

1350 g

Explanation: just add a 0

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Help!!! I don’t know how to solve!!!
ladessa [460]
Just use q=mCDeltaT

q=energy
m=mass
c=specific heat
Delta T= Change in temperature
6 0
3 years ago
Question 2 (2 points)
poizon [28]

Answer:

m_w=439.2g

Explanation:

Hello!

In this case, since the by-mass percent of a solution is a measure of the mass of the solute over the mass of the solution:

\%m/m=\frac{m_{solute}}{m_{solution}} *100\%

As we know the mass of the solution and the by-mass percent, we can compute the mass of glucose in the 480 g of solution:

m_{solute}=\frac{\%m/m*m_{solution}}{100\%}

Thus, by plugging in the data, we obtain:

m_{solute}=\frac{8.5\%*480g}{100\%}=40.8g

Finally, since the solution is made up of glucose and water, we compute the mass of water as follows:

m_w=m_{sol}-m_{solute}=480g-40.8g\\\\m_w=439.2g

Best regards!

7 0
3 years ago
A substance decays so that the amount a of the substance left after t years is given by: a = a 0 · (0.9) t , where a 0 is the or
Dennis_Churaev [7]

I think the correct form of the equation is given as:

a = a0 * (0.9)^t

where t is an exponent of 0.9 since this is an exponential decay of 1st order reaction

 

Now to solve for the half life, this is the time t in which the amount left is half of the original amount, therefore that is when:

a = 0.5 a0

 

Substituting this into the equation:

0.5 a0 = a0 * (0.9)^t

0.5 = (0.9)^t

Taking the log of both sides:

t log 0.9 = log 0.5

t = log 0.5 / log 0.9

t = 6.58 years

 

Answer:

half life = 6.58 years

3 0
3 years ago
A heterogeneous mixture is one in which substances are evenly distributed and have a uniform composition
Sati [7]
If you’re asking if this is true or false, it is false . A HOMOGENEOUS mixture is one the where the substances are evenly distributed and have a uniform composition, while a HETEROGENEOUS mixture is uneven and the substances are part from each like oil and water.
7 0
3 years ago
Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution containing 155 mg of
Cloud [144]

Answer:

The analytical molar concentration is 0,04743M

The equilibrium concentrations are HA = 0,01281M; H⁺ and A⁻ = 0,03462M

Explanation:

The analytical molar concentrations is:

155mg ≡ 0,155g÷163,4g/mol = 9,486x10⁻⁴ moles

9,486x10⁻⁴ moles ÷ 0,020L = <em>0,04743 M</em>

The trichloroacetic acid [HA] dissociates in water in a 73%, thus:

HA ⇄ H⁺ + A⁻

The equilibrium concentrations are:

HA = 0,04743M ×(100-73)% = 0,01281M

H⁺ and A⁻ = 0,04743M ×73% = 0,03462M

I hope it helps!

8 0
4 years ago
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