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mart [117]
3 years ago
8

Tissues combine to form Blank Space __________.

Chemistry
1 answer:
crimeas [40]3 years ago
3 0
Organs because the tissue when form its make organs
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The solubility of nacl at 30°c is 36.3 g/100 g of water. what mass of nacl is contained in a saturated solution that contains 30
sleet_krkn [62]
Because a solubility is calculated for saturated solutions we can write:

36.3 g NaCl ---100g water
x g NaCl ---- 300.0 g water

x=36.3*300/100=108.9 g NaCl
7 0
3 years ago
Read 2 more answers
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N
Rina8888 [55]

Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

7 0
3 years ago
The pH of an acid has nothing to do with the strength of the acid.
Levart [38]

Answer:

false

Explanation:

6 0
3 years ago
How many moles of sodium (Na) are there in a<br> sample of 5.87 x 1024 atoms of sodium?
Leno4ka [110]
9.74x 2351 that's the answer
6 0
3 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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