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3 years ago

Why shouldn’t we add water to sulfuric acid?

Chemistry

1 answer:

Kamila [148]3 years ago

5 0

Answer:

If you add water to sulfuric acid, it can boil and spit and you may get a nasty acid burn.

Explanation:

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PLEASe help. How many ions? With picture.

Likurg_2 [28]

Since this equation is balanced, we know that the law of conversation of mass id applied, and we could calculate easily.

Na= 2
NO3= 2
Ca= 1
<span>Cl= 1</span>

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3 years ago

The time necessary for the decay of one- half sample of a radioactive substance is a _____. mass defect daughter half life nucli

Mazyrski [523]

Answer: half life

Explanation: Radioactive decay follows first order kinetics and the time required for the decay of a radioactive material is calculated as follows:

$t=\frac{2.303}{k}\hspace{1mm}log\frac{x}{a}$

t= time required

k= disintegration constant

x= amount of substance left after time t

a= initial amount of substance

when one half of the sample is decayed, one half of the sample remains and t can be represented as $t_{1/2}$

at $t= t_{1/2}$ , $x=\frac{a}{2}$

$t_{1/2}=\frac{2.303}{k}\hspace{1mm}log\frac{a/2}{a}$

$t_{1/2}=\frac{0.693}{k}$

3 0

3 years ago

Read 2 more answers

What’s the answer to these 3 questions? thanks!

Citrus2011 [14]

Water has h bonding
H-H
Sodium fluoride
I think

4 0

2 years ago

37.9 grams of an unknown substance undergoes a temperature increase of

jenyasd209 [6]

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

4 0

2 years ago

What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer

Zepler [3.9K]

$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$

8 0

3 years ago