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jolli1 [7]
2 years ago
12

What does vascular mean

Chemistry
1 answer:
-BARSIC- [3]2 years ago
5 0
<span>of, relating to, affecting, or consisting of a vessel or vessels, especially those that carry blood</span>
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A teacher places a cup of coffee onto an electronic balance at the front of the science laboratory. The teacher then adds three
motikmotik

Answer:

it will be same as before

4 0
2 years ago
Read 2 more answers
Please help
KonstantinChe [14]

Answer:  There are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given BaSO_{4} solution is as follows.

Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

7 0
3 years ago
Given: 4Na + O2 → 2Na2O In this chemical reaction, how many moles of Na2O will be produced if 2.90 moles of Na react completely?
crimeas [40]

Since we already have the balanced equation, we know that the ratio between Na:Na_{2}O is 4:2 respectively.

So then we can set up a proportion to find the number of moles produced when 2.90 moles of Na react completely:

\frac{4mol_{Na}}{2mol_{Na_{2}O}} =\frac{2.90mol_{Na}}{xmol_{Na_{2}O}}

Then we cross multiply and solve for x:

4x=5.8

x=1.45

Therefore, we know that when 2.90 moles of Na react completely, there are 1.45 moles of Na_{2}O that are produced.

3 0
2 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
I need help with this
Lelechka [254]

Answer:

The answer is its equal to the volume of its container.

--------------------------------------------------------------------------------

I hope this helps! :)

7 0
3 years ago
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